(12x+28)(3x+62)=180

Solution for (12x+28)(3x+62)=180 equation:

(12x+28)(3x+62)=180

We move all terms to the left:

(12x+28)(3x+62)-(180)=0

We multiply parentheses ..

(+36x^2+744x+84x+1736)-180=0

We get rid of parentheses

36x^2+744x+84x+1736-180=0

We add all the numbers together, and all the variables

36x^2+828x+1556=0

a = 36; b = 828; c = +1556;
Δ = b2-4ac
Δ = 8282-4·36·1556
Δ = 461520
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:

$\sqrt{\Delta}=\sqrt{461520}=\sqrt{144*3205}=\sqrt{144}*\sqrt{3205}=12\sqrt{3205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(828)-12\sqrt{3205}}{2*36}=\frac{-828-12\sqrt{3205}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(828)+12\sqrt{3205}}{2*36}=\frac{-828+12\sqrt{3205}}{72}
$