16x^2-8x+(-7)=0

Solution for 16x^2-8x+(-7)=0 equation:

16x^2-8x+(-7)=0

We add all the numbers together, and all the variables

16x^2-8x-7=0

a = 16; b = -8; c = -7;
Δ = b2-4ac
Δ = -82-4·16·(-7)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16\sqrt{2}}{2*16}=\frac{8-16\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16\sqrt{2}}{2*16}=\frac{8+16\sqrt{2}}{32}
$