-16x^2+12x+25=0

Solution for -16x^2+12x+25=0 equation:

-16x^2+12x+25=0

a = -16; b = 12; c = +25;
Δ = b2-4ac
Δ = 122-4·(-16)·25
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{109}}{2*-16}=\frac{-12-4\sqrt{109}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{109}}{2*-16}=\frac{-12+4\sqrt{109}}{-32}
$