2x^2-40x+172=0

Solution for 2x^2-40x+172=0 equation:

2x^2-40x+172=0

a = 2; b = -40; c = +172;
Δ = b2-4ac
Δ = -402-4·2·172
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{14}}{2*2}=\frac{40-4\sqrt{14}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{14}}{2*2}=\frac{40+4\sqrt{14}}{4}
$