How do you solve #x^2+12x-64=0# using the quadratic formula?

Question

4742 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-02T22:10:30

Identify a, b, and c in your equation. Then substitute the values into the quadratic equation. Solve for #x#.

#x^2+12x-64=0#

#a=1;# #b=12;# #c=-64#

#x=(-12+-sqrt(12^2-(4*1*-64)))/(2*1)# =

#x=(-12+-sqrt(144+256))/2# =

#x=(-12+-sqrt(400))/2# =

#x=(-12+-20)/2#

#x=(-12+20)/2# =

#x=8/2# =

#x=4#

#x=(-12-20)/2# =

#x=-32/2# =

#x=-16#

#x=4, -16#