How do you use the quadratic formula to solve $2x^2-5x-2=0$ ?

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Answer 1 · 2017-08-24T17:22:34

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 $color(red)(a)$

$color(blue)(-5)$ for $color(blue)(b)$

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

$x = (-color(blue)((-5)) +- sqrt(color(blue)((-5))^2 - (4 * color(red)(2) * color(green)(-2))))/(2 * color(red)(2))$

$x = (5 +- sqrt(25 - (-16)))/4$

$x = (5 +- sqrt(25 + 16))/4$

$x = (5 +- sqrt(41))/4$

How do you use the quadratic formula to solve 2x^2-5x-2=02x^2-5x-2=0?