How do you solve #x^2+10x=15#?

1 Answer
Apr 6, 2018

Answer:

#x = -5 +- sqrt30#

Here's how I did it:

Explanation:

#x^2 + 10x = 15#

First, we want to set one side to #0# and let one side have #3# terms so that we can factor it, so we subtract #15# from both sides of the equation:
#x^2 + 10x - 15 = 0#

Now we factor. We have to find two numbers that:

  • Multiply up to #-15#
  • Add up to #10#.

We know that the factors of #-15# are #-15, -5, -3, -1, 1, 3, 5,# and #15#. However, no group of factors of #-15# can add up to #10#, so we have to do another method, called the quadratic formula.

The quadratic formula is #x = (-b +- sqrt(b^2 - 4ac))/(2a)#.

Our equation is in the form of #ax^2 + bx^2 + c#, which is also called standard form. So we know that:
#a = 1#

#b = 10#

#c = -15#

Now let's substitute these values into the quadratic formula:
#x = (-10 +- sqrt(10^2 - 4(1)(-5)))/(2(1))#

Simplify by doing #10^2#, #-4(1)(-5)#, and #2(1)#:
#x = (-10 +- sqrt(100 + 20))/2#

Add #100 + 20#:
#x = (-10 +- sqrt(120))/2#

Radicalize/simplify #120#
#x = (-10 +- sqrt(4*30))/2#

#x = (-10 +- 2sqrt30)/2#

Divide by #2#:
#x = -5 +- sqrt30#

Hope this helps!