16x^2+16x-25=0

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

16x^2+16x-25=0

a = 16; b = 16; c = -25;
Δ = b2-4ac
Δ = 162-4·16·(-25)
Δ = 1856
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:

$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{29}}{2*16}=\frac{-16-8\sqrt{29}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{29}}{2*16}=\frac{-16+8\sqrt{29}}{32}
$