16x^2-8x+(-6)=0

Solution for 16x^2-8x+(-6)=0 equation:

16x^2-8x+(-6)=0

We add all the numbers together, and all the variables

16x^2-8x-6=0

a = 16; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·16·(-6)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{7}}{2*16}=\frac{8-8\sqrt{7}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{7}}{2*16}=\frac{8+8\sqrt{7}}{32}
$