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

1 Answer
Aug 6, 2017

Answer:

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

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#