x^2-12x+25=40

Solution for x^2-12x+25=40 equation:

x^2-12x+25=40

We move all terms to the left:

x^2-12x+25-(40)=0

We add all the numbers together, and all the variables

x^2-12x-15=0

a = 1; b = -12; c = -15;
Δ = b2-4ac
Δ = -122-4·1·(-15)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{51}}{2*1}=\frac{12-2\sqrt{51}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{51}}{2*1}=\frac{12+2\sqrt{51}}{2}
$