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What are the solutions to this equation?

[tex]\[ 2x^2 = -10x + 12 \][/tex]

A. [tex]\( x = -6 \)[/tex]
B. [tex]\( x = 3 \)[/tex]
C. [tex]\( x = -2 \)[/tex]
D. [tex]\( x = 6 \)[/tex]
E. [tex]\( x = 1 \)[/tex]
F. [tex]\( x = -3 \)[/tex]


Sagot :

To solve the quadratic equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex], we can use the quadratic formula, which is given by:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, the coefficients are:
[tex]\[ a = 2, \quad b = 10, \quad c = -12 \][/tex]

First, we calculate the discriminant:

[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 10^2 - 4 \cdot 2 \cdot (-12) \][/tex]
[tex]\[ \Delta = 100 + 96 \][/tex]
[tex]\[ \Delta = 196 \][/tex]

Since the discriminant [tex]\(\Delta\)[/tex] is positive, there are two distinct real solutions. We now use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{196}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-10 \pm 14}{4} \][/tex]

This gives us two solutions:

1. [tex]\[ x_1 = \frac{-10 + 14}{4} = \frac{4}{4} = 1 \][/tex]

2. [tex]\[ x_2 = \frac{-10 - 14}{4} = \frac{-24}{4} = -6 \][/tex]

Thus, the solutions to the equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex] are:

[tex]\[ x = 1 \quad \text{and} \quad x = -6 \][/tex]

Therefore, out of the provided options, the correct answers are:
- [tex]\(x = -6\)[/tex]
- [tex]\(x = 1\)[/tex]
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