16x^2-8x+(-8)=0

Solution for 16x^2-8x+(-8)=0 equation:

16x^2-8x+(-8)=0

We add all the numbers together, and all the variables

16x^2-8x-8=0

a = 16; b = -8; c = -8;
Δ = b2-4ac
Δ = -82-4·16·(-8)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24}{2*16}=\frac{-16}{32}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*16}=\frac{32}{32}
=1
$