Use the quadratic formula, [tex]$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$[/tex], to solve the equation [tex]$2x^2 + 10x - 6 = 0$[/tex]. Round to the nearest hundredths place.

A. [tex][tex]$x = -5.54$[/tex][/tex] and [tex]$x = 0.54$[/tex]
B. [tex]$x = -0.54$[/tex] and [tex][tex]$x = 5.54$[/tex][/tex]
C. [tex]$x = -4.30$[/tex] and [tex]$x = -0.70$[/tex]
D. [tex][tex]$x = -22.17$[/tex][/tex] and [tex]$x = 2.17$[/tex]



Answer :

To solve the quadratic equation [tex]\(2x^2 + 10x - 6 = 0\)[/tex] using the quadratic formula, we follow these steps:

1. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
The quadratic equation is in the form [tex]\(ax^2 + bx + c = 0\)[/tex].
[tex]\[ a = 2, \quad b = 10, \quad c = -6 \][/tex]

2. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
The discriminant is given by the formula [tex]\(\Delta = b^2 - 4ac\)[/tex].
[tex]\[ \Delta = 10^2 - 4 \times 2 \times (-6) \][/tex]
[tex]\[ \Delta = 100 + 48 \][/tex]
[tex]\[ \Delta = 148 \][/tex]

3. Apply the quadratic formula:
The roots are given by the formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values we have:
[tex]\[ x = \frac{-10 \pm \sqrt{148}}{2 \times 2} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{148}}{4} \][/tex]

4. Calculate the two roots:

[tex]\[ x_1 = \frac{-10 + \sqrt{148}}{4} \][/tex]
[tex]\[ x_2 = \frac{-10 - \sqrt{148}}{4} \][/tex]

Using a calculator to find [tex]\(\sqrt{148} \approx 12.17\)[/tex], we then find:
[tex]\[ x_1 = \frac{-10 + 12.17}{4} \approx \frac{2.17}{4} \approx 0.54 \][/tex]
[tex]\[ x_2 = \frac{-10 - 12.17}{4} \approx \frac{-22.17}{4} \approx -5.54 \][/tex]

5. Round the results to the nearest hundredths place:
[tex]\[ x_1 \approx 0.54 \][/tex]
[tex]\[ x_2 \approx -5.54 \][/tex]

6. Conclusion:
The solutions to the equation [tex]\(2x^2 + 10x - 6 = 0\)[/tex] are:
[tex]\[ x = 0.54 \quad \text{and} \quad x = -5.54 \][/tex]

Therefore, the correct answer from the given options is:
[tex]\[ \boxed{-5.54 \text{ and } 0.54} \][/tex]