w^2-32w+152=0

Solution for w^2-32w+152=0 equation:

w^2-32w+152=0

a = 1; b = -32; c = +152;
Δ = b2-4ac
Δ = -322-4·1·152
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{26}}{2*1}=\frac{32-4\sqrt{26}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{26}}{2*1}=\frac{32+4\sqrt{26}}{2}
$