90+(3x^2+60)+(2x^2-25x)=180

Solution for 90+(3x^2+60)+(2x^2-25x)=180 equation:

90+(3x^2+60)+(2x^2-25x)=180

We move all terms to the left:

90+(3x^2+60)+(2x^2-25x)-(180)=0

We add all the numbers together, and all the variables

(3x^2+60)+(2x^2-25x)-90=0

We get rid of parentheses

3x^2+2x^2-25x+60-90=0

We add all the numbers together, and all the variables

5x^2-25x-30=0

a = 5; b = -25; c = -30;
Δ = b2-4ac
Δ = -252-4·5·(-30)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-35}{2*5}=\frac{-10}{10}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+35}{2*5}=\frac{60}{10}
=6
$