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

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Answer 1 · 2017-08-28T12:28:06

See a solution process below:

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

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

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

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

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

$x = (-color(blue)((-3)) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(2) * color(green)(-6))))/(2 * color(red)(2))$

$x = (color(blue)(3) +- sqrt(9 - (-48)))/4$

$x = (color(blue)(3) +- sqrt(9 + 48))/4$

$x = (color(blue)(3) +- sqrt(57))/4$

How do you solve 2x^2-3x-6=02x^2-3x-6=0 using the quadratic formula?