x^2+20=125

Solution for x^2+20=125 equation:

x^2+20=125

We move all terms to the left:

x^2+20-(125)=0

We add all the numbers together, and all the variables

x^2-105=0

a = 1; b = 0; c = -105;
Δ = b2-4ac
Δ = 02-4·1·(-105)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*1}=\frac{0-2\sqrt{105}}{2}
=-\frac{2\sqrt{105}}{2}
=-\sqrt{105}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*1}=\frac{0+2\sqrt{105}}{2}
=\frac{2\sqrt{105}}{2}
=\sqrt{105}
$