4x^2+4=26

Solution for 4x^2+4=26 equation:

4x^2+4=26

We move all terms to the left:

4x^2+4-(26)=0

We add all the numbers together, and all the variables

4x^2-22=0

a = 4; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·4·(-22)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{22}}{2*4}=\frac{0-4\sqrt{22}}{8}
=-\frac{4\sqrt{22}}{8}
=-\frac{\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{22}}{2*4}=\frac{0+4\sqrt{22}}{8}
=\frac{4\sqrt{22}}{8}
=\frac{\sqrt{22}}{2}
$