(4x^2-21)+(3x^2-4x)=180

Solution for (4x^2-21)+(3x^2-4x)=180 equation:

(4x^2-21)+(3x^2-4x)=180

We move all terms to the left:

(4x^2-21)+(3x^2-4x)-(180)=0

We get rid of parentheses

4x^2+3x^2-4x-21-180=0

We add all the numbers together, and all the variables

7x^2-4x-201=0

a = 7; b = -4; c = -201;
Δ = b2-4ac
Δ = -42-4·7·(-201)
Δ = 5644
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:

$\sqrt{\Delta}=\sqrt{5644}=\sqrt{4*1411}=\sqrt{4}*\sqrt{1411}=2\sqrt{1411}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{1411}}{2*7}=\frac{4-2\sqrt{1411}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{1411}}{2*7}=\frac{4+2\sqrt{1411}}{14}
$