4(3^2x-5)=80

Solution for 4(3^2x-5)=80 equation:

4(3^2x-5)=80

We move all terms to the left:

4(3^2x-5)-(80)=0

We multiply parentheses

12x^2-20-80=0

We add all the numbers together, and all the variables

12x^2-100=0

a = 12; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·12·(-100)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*12}=\frac{0-40\sqrt{3}}{24}
=-\frac{40\sqrt{3}}{24}
=-\frac{5\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*12}=\frac{0+40\sqrt{3}}{24}
=\frac{40\sqrt{3}}{24}
=\frac{5\sqrt{3}}{3}
$