(4x+5)+(x^2-27)=180

Solution for (4x+5)+(x^2-27)=180 equation:

(4x+5)+(x^2-27)=180

We move all terms to the left:

(4x+5)+(x^2-27)-(180)=0

We get rid of parentheses

x^2+4x+5-27-180=0

We add all the numbers together, and all the variables

x^2+4x-202=0

a = 1; b = 4; c = -202;
Δ = b2-4ac
Δ = 42-4·1·(-202)
Δ = 824
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:

$\sqrt{\Delta}=\sqrt{824}=\sqrt{4*206}=\sqrt{4}*\sqrt{206}=2\sqrt{206}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{206}}{2*1}=\frac{-4-2\sqrt{206}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{206}}{2*1}=\frac{-4+2\sqrt{206}}{2}
$