16x^2-116x+96=0

Solution for 16x^2-116x+96=0 equation:

16x^2-116x+96=0

a = 16; b = -116; c = +96;
Δ = b2-4ac
Δ = -1162-4·16·96
Δ = 7312
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:

$\sqrt{\Delta}=\sqrt{7312}=\sqrt{16*457}=\sqrt{16}*\sqrt{457}=4\sqrt{457}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-116)-4\sqrt{457}}{2*16}=\frac{116-4\sqrt{457}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-116)+4\sqrt{457}}{2*16}=\frac{116+4\sqrt{457}}{32}
$