x^2-40x+12=0

Solution for x^2-40x+12=0 equation:

x^2-40x+12=0

a = 1; b = -40; c = +12;
Δ = b2-4ac
Δ = -402-4·1·12
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{97}}{2*1}=\frac{40-4\sqrt{97}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{97}}{2*1}=\frac{40+4\sqrt{97}}{2}
$