16x^2-132x+28=0

Solution for 16x^2-132x+28=0 equation:

16x^2-132x+28=0

a = 16; b = -132; c = +28;
Δ = b2-4ac
Δ = -1322-4·16·28
Δ = 15632
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:

$\sqrt{\Delta}=\sqrt{15632}=\sqrt{16*977}=\sqrt{16}*\sqrt{977}=4\sqrt{977}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-132)-4\sqrt{977}}{2*16}=\frac{132-4\sqrt{977}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-132)+4\sqrt{977}}{2*16}=\frac{132+4\sqrt{977}}{32}
$