4(2^2x-1)+8=48

Solution for 4(2^2x-1)+8=48 equation:

4(2^2x-1)+8=48

We move all terms to the left:

4(2^2x-1)+8-(48)=0

We add all the numbers together, and all the variables

4(2^2x-1)-40=0

We multiply parentheses

8x^2-4-40=0

We add all the numbers together, and all the variables

8x^2-44=0

a = 8; b = 0; c = -44;
Δ = b2-4ac
Δ = 02-4·8·(-44)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{22}}{2*8}=\frac{0-8\sqrt{22}}{16}
=-\frac{8\sqrt{22}}{16}
=-\frac{\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{22}}{2*8}=\frac{0+8\sqrt{22}}{16}
=\frac{8\sqrt{22}}{16}
=\frac{\sqrt{22}}{2}
$