3(2^2x+3)-19=8

Solution for 3(2^2x+3)-19=8 equation:

3(2^2x+3)-19=8

We move all terms to the left:

3(2^2x+3)-19-(8)=0

We add all the numbers together, and all the variables

3(2^2x+3)-27=0

We multiply parentheses

6x^2+9-27=0

We add all the numbers together, and all the variables

6x^2-18=0

a = 6; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·6·(-18)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*6}=\frac{0-12\sqrt{3}}{12}
=-\frac{12\sqrt{3}}{12}
=-\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*6}=\frac{0+12\sqrt{3}}{12}
=\frac{12\sqrt{3}}{12}
=\sqrt{3}
$