16x^2-160x+168=0

Solution for 16x^2-160x+168=0 equation:

16x^2-160x+168=0

a = 16; b = -160; c = +168;
Δ = b2-4ac
Δ = -1602-4·16·168
Δ = 14848
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:

$\sqrt{\Delta}=\sqrt{14848}=\sqrt{256*58}=\sqrt{256}*\sqrt{58}=16\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16\sqrt{58}}{2*16}=\frac{160-16\sqrt{58}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16\sqrt{58}}{2*16}=\frac{160+16\sqrt{58}}{32}
$