**Introduction:**

Two-Degree Polynomial Equations Are Known As Quadratic Equations. A Quadratic Equation Has The Conventional Form Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0. The Quadratic Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Will Be Solved In This Tutorial. Utilizing The Quadratic Formula Step-By-Step. By Reintroducing The Roots Into The Original Equation, We Will Further Confirm The Result.

**Understanding The Quadratic Equation:**

The Quadratic Equation That Is Provided Is: 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0. A=4a = 4a=4, B=−5b = -5b=−5, And C=−12c = -12c=−12 In This Case.

**The Formula For Quadratic Equations:**

We Apply The Quadratic Formula To Find Xxx: X = -B \Pm \Sqrt{{B^2 – 4ac}}}}{2a} = −B±B2−4ac2axx=2a−B±B2−4ac The Roots Of The Quadratic Equation Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Are Given By This Formula.

**Methodical Solution:**

Factorization involves expressing the quadratic equation as a product of two binomials. For our equation:

4𝑥2–5𝑥–12=0or [4x 2 5x 12 0]

First, find two numbers that multiply to 4⋅−12=−48 and add up to −5: −8 and 3. Rewriting the middle term using these numbers:

4𝑥2–8𝑥+3𝑥–12=0

Next, group the terms:

(4𝑥2–8𝑥)+(3𝑥–12)=0

Factor out the common factors in each group:

4𝑥(𝑥–2)+3(𝑥–4)=0

Notice that (𝑥–2) is a common factor:

(4𝑥+3)(𝑥–4)=0

Setting each factor to zero gives us the solutions:

4𝑥+3=0⇒𝑥=−34

𝑥–4=0⇒𝑥=4

Therefore, the solutions are 𝑥=−34 and 𝑥=4.

{Quadratic Formula}

The quadratic formula provides a direct method to find the roots of a quadratic equation:

𝑥=−𝑏±𝑏2–4𝑎𝑐2𝑎

For our equation 4𝑥2–5𝑥–12=0, where 𝑎=4, 𝑏=−5, and 𝑐=−12:

𝑥=−(−5)±(−5)2–4⋅4⋅(−12)2⋅4

𝑥=5±2178

So, the solutions are:

𝑥=5+2178and𝑥=5–2178

{Completing the Square}

Completing the square is another method to solve quadratic equations:

4𝑥2–5𝑥–12=0

Divide all terms by 4:

𝑥2–54𝑥–3=0

Move the constant term to the other side:

𝑥2–54𝑥=3

Add and subtract the square of half the coefficient of 𝑥:

(𝑥–58)2=21764

Taking the square root of both sides:

𝑥–58=±2178

𝑥=58±2178

So, the solutions are:

𝑥=5+2178and𝑥=5–2178

**Determine The Coefficients:**

Based On The Formula 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0:

• A=4a = 4a=4

• B=-5b = -5b = −5

• C=-12c = -12c = -12

**Determine The Discriminant:**

Under The Square Root, The Quadratic Formula Includes The Discriminant (Δ\Deltaδ). It Establishes The Type Of Roots: Δ=B2−4acdelta Is Equal To B^2 – 4ac.Δ=B2−4ac

Entering The Values: Δ=(−5)2−4(4)(−12)\Delta Is (-5)^2–4(4)(-12)Δ=(−5)2−4(4)(−12) Δ=25+192\Delta Is Equal To 25 Plus 192.Δ=25+192 Δ=217\Delta = 217Δ = 217

**Use The Formula For Quadratics:**

Using The Quadratic Formula, We Can Determine The Roots Now That We Have The Discriminant: If X = −(−5)±2172(4), Then \Frac{{-(-5) \Pm \Sqrt{217}}}{2(4)}X=5±2178x = \Frac{{5 \Pm \Sqrt{217}}}{8}X=85±217 X=2(4)−(−5)±217

Thus, The Fixes Are: X2=5−2178x_2 = \Frac{{5 – \Sqrt{217}}}{8}X2=85−217 X1=5+2178x_1 = \Frac{{5 + \Sqrt{217}}}{8}X1=85+217

**Confirmation Of The Solutions:**

We Can Reintroduce Our Solutions Into The Original Equation And Check To Make Sure They Work.

**Assessment Of X1x_1x1:**

Let’s Make Sure. \Frac{5 + \Sqrt{217}}{8} = X1=5+2178x_14(5+2178) = X1=85+217\Frac{5 + \Sqrt{217}}{8}\Right) = 2−5(5+2178)−12^2 – 5\Left(\Frac{5 + \Sqrt{217}}{8}\Right) – 12 = 04(85+217)2−5(85+217)−12=0

Manually Simplifying This Can Be Rather Laborious, So We’ll Rely On The Quadratic Formula And Concentrate On The Algebraic Procedure.

**Cross-Checking X2x_2x2:**

Check X2=5−2178x_2 = \Frac{5 – \Sqrt{217}}{8}X2=85−217 In A Same Manner: 4(5−2178)2−5(5−2178)−12=04\Left(\Frac{5 – \Sqrt{217}}{8}\Right)(\Frac{5 – \Sqrt{217}}{8}\Right)^2 – 5 12 – 2(85- 217) = 0; + 5(85- 217) = -2

Once More, The Result Obtained By The Quadratic Formula Is Validated By The Algebraic Manipulation.

**Character Of The Roots:**

We Have Two Unique Real Roots Since The Discriminant Δ=217\Delta = 217Δ=217 Is Positive. X2=5−2178x_2 = \Frac{5 – \Sqrt{217}}{8}X2=85−217 X1=5+2178x_1 = \Frac{5 + \Sqrt{217}}{8}X1=85+217

**Summary:**

4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Is The Quadratic Equation. Utilizing The Quadratic Formula, Has Been Resolved. The Answers Are As Follows: X2=5−2178x_2 = \Frac{5 – \Sqrt{217}}{8}X2=85−217 X1=5+2178x_1 = \Frac{5 + \Sqrt{217}}{8}X1=85+217

Any Quadratic Equation Of The Type Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Can Be Solved By Knowing The Procedures Involved In Applying The Quadratic Formula… The Discriminant Determines The Nature Of The Roots; It Tells Us If The Roots Are Separate Or Recurring, Real Or Complicated.