(4x^2-36)+(3x^2-5x)=180

Solution for (4x^2-36)+(3x^2-5x)=180 equation:

(4x^2-36)+(3x^2-5x)=180

We move all terms to the left:

(4x^2-36)+(3x^2-5x)-(180)=0

We get rid of parentheses

4x^2+3x^2-5x-36-180=0

We add all the numbers together, and all the variables

7x^2-5x-216=0

a = 7; b = -5; c = -216;
Δ = b2-4ac
Δ = -52-4·7·(-216)
Δ = 6073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{6073}}{2*7}=\frac{5-\sqrt{6073}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{6073}}{2*7}=\frac{5+\sqrt{6073}}{14}
$