(3x^2+10)+(9x+50)=180

Solution for (3x^2+10)+(9x+50)=180 equation:

(3x^2+10)+(9x+50)=180

We move all terms to the left:

(3x^2+10)+(9x+50)-(180)=0

We get rid of parentheses

3x^2+9x+10+50-180=0

We add all the numbers together, and all the variables

3x^2+9x-120=0

a = 3; b = 9; c = -120;
Δ = b2-4ac
Δ = 92-4·3·(-120)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-39}{2*3}=\frac{-48}{6}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+39}{2*3}=\frac{30}{6}
=5
$