how to do this quadratic equation X² + 12x – 64 = 0?

x = 4

x = -16

good luck

wow... I used to do these blindfolded at school! I'm shocked at myself that I forgot it all! Hope some one has the answer!

( x + 16 ) ( x - 4 ) = 0

x = - 16 , x = 4

x^2 + 12x - 64 = 0
x^2 + 16x - 4x - 64 = 0
(x^2 + 16x) - (4x + 64) = 0
x(x + 16) - 4(x + 16) = 0
(x + 16)(x - 4) = 0

x + 16 = 0
x = -16

x - 4 = 0
x = 4

∴ x = -16, 4

you factor it. by finding two numbers that have a product of -64 and a sum of 12 you get:
(x-4)(x+16)
then whatever two numbers make this equal zero are your x intercepts and solutions or squares
they are 4 and -16

x² + 12x – 64 = 0
factors into:
(x + 16)(x - 4) = 0
set each to 0
x + 16 = 0 and x - 4 = 0
solve for x
x = -16 and x = 4

X² + 12x – 64 = 0
(x+16)(x-4)=0
| |
x+16=0 x-4=0
x=-16 x=4

Check:

X²+12x–64=0
(-16)²+12(-16)-64=0
256-192-64=0
64-64=0
0=0√

X²+12x–64=0
(4)²+12(4)-64=0
16+48-64=0
64-64=0
0=0√

Factor.
(x + 16)(x - 4) = 0
x = -16, 4

x*2 +16x-4x-64=0
x(x+16)-4(x+16)=0
(x-4)(x+16)=0
x=4,-16

X² + 12x – 64 = 0
(x + 16)(x - 4) = 0
x + 16 = 0
x = - 16
x - 4 = 0
x = 4


Make sense? Good luck.