How do you solve #x^2 - 3x = 40#?

1 Answer
Feb 16, 2016

Answer:

#x=8,-5#

Explanation:

#color(blue)(x^2-3x=40#

Subtract #40# both sides

We get:

#color(red)(x^2-3x-40=0#

You can solve this both by Factoring and using the Quadratic formula:

#1)# First we can solve it by factoring:

#color(red)(x^2-3x-40=0#

Factor #x^2-3x-40#

We get

#rarr(x-8)(x+5)=0#

If you solve it you get

#rArrcolor(green)(x=8,-5#

#2)# Quadratic formula

#color(red)(x^2-3x-40=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

So use Quadratic formula:

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case #a=1,b=-3,c=-40#

Substitute the values into the equation:

#rarrx=(-(-3)+-sqrt(-3^2-4(1)(-40)))/(2(1))#

#rarrx=(3+-sqrt(9-(-160)))/(2)#

#rarrx=(3+-sqrt(9+160))/(2)#

#rarrx=(3+-sqrt(169))/(2)#

#rArrx=color(indigo)(3+-13)/2#

Now we have two solutions:

#x=color(orange)((3+13)/2),color(violet)((3-13)/2#

Solve for the first and then into the second:

#rarrx=color(orange)((3+13)/2#

#rArrcolor(green)(x=16/2=8#

For the second

#rarrcolor(violet)(x=(3-13)/2#

#rArrcolor(green)(x=-10/2=-5#