How do you solve $0= -16t^2+5t+30$ ?

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Answer 1 · 2017-08-06T11:21:24

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:

$color(red)(-16)$ for $color(red)(a)$

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

$color(green)(30)$ for $color(green)(c)$ gives:

$x = (-color(blue)(5) +- sqrt(color(blue)(5)^2 - (4 * color(red)(-16) * color(green)(30))))/(2 * color(red)(-16))$

$x = (-color(blue)(5) +- sqrt(25 - (-1920)))/-32$

$x = (-color(blue)(5) +- sqrt(25 + 1920))/-32$

$x = (-color(blue)(5) +- sqrt(1945))/-32$

$x = (color(blue)(5) +- sqrt(1945))/32$

How do you solve 0= -16t^2+5t+30 0= -16t^2+5t+30 ?