How do you solve x^2+5X=150?

Question

I'm doing one of those summer review packets...and I forgot how to solve this basic problem.

Answer 1

x^2 + 5x - 150 = 0
x^2 + 15x - 10x - 150 = 0
x(x + 15) - 10(x + 15) = 0
(x + 15)(x - 10) = 0

x = -15, x = 10

suppose, ax^2 + bx + c is the general eq...just split 'b' into two numbers that add upto 'b' and multiply to give the product of 'a' and 'c'.

Answer 2

First rearrange the problem.

x^2 + 5x - 150 = 0
Then factor
x^2 + 5x - 150 = (x - 10)(x + 15) = 0
So either:
x - 10 = 0, and x = 10
or x + 15 = 0, and x = -15

If you're not very good at factoring, you can always use the quadratic equation:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
In this case,
x = [-5 +- sqrt(5^2 - 4(-150))]/2
x = [-5 +- sqrt(625)]/2
x = [-5 +- 25]/2
x = 10 or x = -15

Answer 3

x² + 5x - 150 = 0
(x - 10) (x + 15) = 0
x = 10 , x = - 15

Answer 4

complete the square

x^2 + 5x + 25/4 = 150 + 25/4
(x + 5/2)^2 = 150 + 25/4
x+ 5/2 = 12.5
x = 10

Answer 5

x^2+5x-150=0
(x+15)(x-10)=0
x+15=0 or x-10=0
x=-15 x=10

Answer 6

x^2+5X=150
x^2+5x-150 =0
x^2 +10x - 15x -150 =0
x(x + 10) -15(x + 10) =0
(x - 15)(x+10) =0
x= 15
or
x = -10