How do you solve $2 x^{2} - 3 x - 6 = 0$ $2x^2-3x-6=0$ using the quadratic formula?

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How do you solve $2 x^{2} - 3 x - 6 = 0$ $2x^2-3x-6=0$ using the quadratic formula?

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Answer 1 · 2017-08-28T12:28:06

See a solution process below:

Explanation:

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$2$ $color(red)(2)$ for $a$ $color(red)(a)$

$- 3$ $color(blue)(-3)$ for $b$ $color(blue)(b)$

$- 6$ $color(green)(-6)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - (4 \cdot 2 \cdot - 6)}}{2 \cdot 2}$ $x = (-color(blue)((-3)) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(2) * color(green)(-6))))/(2 * color(red)(2))$

$x = \frac{3 \pm \sqrt{9 - (- 48)}}{4}$ $x = (color(blue)(3) +- sqrt(9 - (-48)))/4$

$x = \frac{3 \pm \sqrt{9 + 48}}{4}$ $x = (color(blue)(3) +- sqrt(9 + 48))/4$

$x = \frac{3 \pm \sqrt{57}}{4}$ $x = (color(blue)(3) +- sqrt(57))/4$

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How do you solve $2 x^{2} - 3 x - 6 = 0$ $2x^2-3x-6=0$ using the quadratic formula?

1 Answer

Explanation:

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