How do you solve $-x^2+x+1=0$ using the quadratic formula?

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Answer 1 · 2017-08-24T18:16:40

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:

$color(red)(-1)$ for $color(red)(a)$

$color(blue)(1)$ for $color(blue)(b)$

$color(green)(1)$ for $color(green)(c)$ gives:

$x = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(-1) * color(green)(1))))/(2 * color(red)(-1))$

$x = (-color(blue)(1) +- sqrt(1 - (-4)))/(-2)$

$x = (-color(blue)(1) +- sqrt(1 + 4))/(-2)$

$x = (-color(blue)(1) +- sqrt(5))/(-2)$

$x = (color(blue)(1) +- sqrt(5))/(2)$

Or

$1/2 +- sqrt(5)/2$

How do you solve -x^2+x+1=0-x^2+x+1=0 using the quadratic formula?