3(2^2x+3)-5(2^x+2)-156=0

Solution for 3(2^2x+3)-5(2^x+2)-156=0 equation:

3(2^2x+3)-5(2^x+2)-156=0

We multiply parentheses

6x^2-10x+9-10-156=0

We add all the numbers together, and all the variables

6x^2-10x-157=0

a = 6; b = -10; c = -157;
Δ = b2-4ac
Δ = -102-4·6·(-157)
Δ = 3868
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:

$\sqrt{\Delta}=\sqrt{3868}=\sqrt{4*967}=\sqrt{4}*\sqrt{967}=2\sqrt{967}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{967}}{2*6}=\frac{10-2\sqrt{967}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{967}}{2*6}=\frac{10+2\sqrt{967}}{12}
$