16x^2+16x-36=0

Solution for 16x^2+16x-36=0 equation:

16x^2+16x-36=0

a = 16; b = 16; c = -36;
Δ = b2-4ac
Δ = 162-4·16·(-36)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{10}}{2*16}=\frac{-16-16\sqrt{10}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{10}}{2*16}=\frac{-16+16\sqrt{10}}{32}
$