-16x^2+16x+23=0

Solution for -16x^2+16x+23=0 equation:

-16x^2+16x+23=0

a = -16; b = 16; c = +23;
Δ = b2-4ac
Δ = 162-4·(-16)·23
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24\sqrt{3}}{2*-16}=\frac{-16-24\sqrt{3}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24\sqrt{3}}{2*-16}=\frac{-16+24\sqrt{3}}{-32}
$