(4x^2-10)+(8x+5)+(x^2+2x+10)=180

Solution for (4x^2-10)+(8x+5)+(x^2+2x+10)=180 equation:

(4x^2-10)+(8x+5)+(x^2+2x+10)=180

We move all terms to the left:

(4x^2-10)+(8x+5)+(x^2+2x+10)-(180)=0

We get rid of parentheses

4x^2+x^2+8x+2x-10+5+10-180=0

We add all the numbers together, and all the variables

5x^2+10x-175=0

a = 5; b = 10; c = -175;
Δ = b2-4ac
Δ = 102-4·5·(-175)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-60}{2*5}=\frac{-70}{10}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+60}{2*5}=\frac{50}{10}
=5
$