-16x^2+16x+6=0

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

-16x^2+16x+6=0

a = -16; b = 16; c = +6;
Δ = b2-4ac
Δ = 162-4·(-16)·6
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{10}}{2*-16}=\frac{-16-8\sqrt{10}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{10}}{2*-16}=\frac{-16+8\sqrt{10}}{-32}
$