(3x+36)(2x+42)=1440

Solution for (3x+36)(2x+42)=1440 equation:

(3x+36)(2x+42)=1440

We move all terms to the left:

(3x+36)(2x+42)-(1440)=0

We multiply parentheses ..

(+6x^2+126x+72x+1512)-1440=0

We get rid of parentheses

6x^2+126x+72x+1512-1440=0

We add all the numbers together, and all the variables

6x^2+198x+72=0

a = 6; b = 198; c = +72;
Δ = b2-4ac
Δ = 1982-4·6·72
Δ = 37476
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:

$\sqrt{\Delta}=\sqrt{37476}=\sqrt{36*1041}=\sqrt{36}*\sqrt{1041}=6\sqrt{1041}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(198)-6\sqrt{1041}}{2*6}=\frac{-198-6\sqrt{1041}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(198)+6\sqrt{1041}}{2*6}=\frac{-198+6\sqrt{1041}}{12}
$