x^2+x^2+22x+121=4x^2

Solution for x^2+x^2+22x+121=4x^2 equation:

x^2+x^2+22x+121=4x^2

We move all terms to the left:

x^2+x^2+22x+121-(4x^2)=0

determiningTheFunctionDomain
x^2+x^2-4x^2+22x+121=0

We add all the numbers together, and all the variables

-2x^2+22x+121=0

a = -2; b = 22; c = +121;
Δ = b2-4ac
Δ = 222-4·(-2)·121
Δ = 1452
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:

$\sqrt{\Delta}=\sqrt{1452}=\sqrt{484*3}=\sqrt{484}*\sqrt{3}=22\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22\sqrt{3}}{2*-2}=\frac{-22-22\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22\sqrt{3}}{2*-2}=\frac{-22+22\sqrt{3}}{-4}
$