How do you solve $2p^2 - 3p -2 = 0$ ?

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Answer 1 · 2018-03-07T09:09:19

See a solution process below:

We can use the quadratic equation to solve this problem:

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 $color(blue)(b)$

$color(green)(-2)$ for $color(green)(c)$ gives:

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

$p = (color(blue)(3) +- sqrt(9 - (-16)))/4$

$p = (color(blue)(3) +- sqrt(9 + 16))/4$

$p = (color(blue)(3) +- sqrt(25))/4$

$p = (color(blue)(3) - 5)/4$ ; $p = (color(blue)(3) + 5)/4$

$p = -2/4$ ; $p = 8/4$

$p = -1/2$ ; $p = 2$

How do you solve 2p^2 - 3p -2 = 02p^2 - 3p -2 = 0?