(10x-4)+(x^2+4)=180

Solution for (10x-4)+(x^2+4)=180 equation:

(10x-4)+(x^2+4)=180

We move all terms to the left:

(10x-4)+(x^2+4)-(180)=0

We get rid of parentheses

x^2+10x-4+4-180=0

We add all the numbers together, and all the variables

x^2+10x-180=0

a = 1; b = 10; c = -180;
Δ = b2-4ac
Δ = 102-4·1·(-180)
Δ = 820
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:

$\sqrt{\Delta}=\sqrt{820}=\sqrt{4*205}=\sqrt{4}*\sqrt{205}=2\sqrt{205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{205}}{2*1}=\frac{-10-2\sqrt{205}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{205}}{2*1}=\frac{-10+2\sqrt{205}}{2}
$