4x^2+20=124

Solution for 4x^2+20=124 equation:

4x^2+20=124

We move all terms to the left:

4x^2+20-(124)=0

We add all the numbers together, and all the variables

4x^2-104=0

a = 4; b = 0; c = -104;
Δ = b2-4ac
Δ = 02-4·4·(-104)
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{26}}{2*4}=\frac{0-8\sqrt{26}}{8}
=-\frac{8\sqrt{26}}{8}
=-\sqrt{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{26}}{2*4}=\frac{0+8\sqrt{26}}{8}
=\frac{8\sqrt{26}}{8}
=\sqrt{26}
$