How do you find the zeros of $y = -x^2 +32x + 19$ using the quadratic formula?

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Answer 1 · 2017-08-29T20:22:47

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 $color(red)(a)$

$color(blue)(32)$ for $color(blue)(b)$

$color(green)(19)$ for $color(green)(c)$ gives:

$x = (-color(blue)(32) +- sqrt(color(blue)(32)^2 - (4 * color(red)(-1) * color(green)(19))))/(2 * color(red)(-1))$

$x = (-color(blue)(32) +- sqrt(1024 - (-76)))/-2$

$x = (-color(blue)(32) +- sqrt(1024 + 76))/-2$

$x = (-color(blue)(32) +- sqrt(1100))/-2$

$x = (-color(blue)(32) - sqrt(100 * 11))/-2$ and $x = (-color(blue)(32) + sqrt(100 * 11))/-2$

$x = (-color(blue)(32) - sqrt(100)sqrt(11))/-2$ and $x = (-color(blue)(32) + sqrt(100)sqrt(11))/-2$

$x = (-color(blue)(32) - 10sqrt(11))/-2$ and $x = (-color(blue)(32) + 10sqrt(11))/-2$

$x = (-color(blue)(32))/-2 - (10sqrt(11))/-2$ and $x = (-color(blue)(32))/-2 + (10sqrt(11))/-2$

$x = 16 + 5sqrt(11)$ and $x = 16 - 5sqrt(11)$

How do you find the zeros of y = -x^2 +32x + 19 y = -x^2 +32x + 19 using the quadratic formula?