(x^2+4x)+(21x+27)=180

Solution for (x^2+4x)+(21x+27)=180 equation:

(x^2+4x)+(21x+27)=180

We move all terms to the left:

(x^2+4x)+(21x+27)-(180)=0

We get rid of parentheses

x^2+4x+21x+27-180=0

We add all the numbers together, and all the variables

x^2+25x-153=0

a = 1; b = 25; c = -153;
Δ = b2-4ac
Δ = 252-4·1·(-153)
Δ = 1237
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{-(25)-\sqrt{1237}}{2*1}=\frac{-25-\sqrt{1237}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{1237}}{2*1}=\frac{-25+\sqrt{1237}}{2}
$