(2x^2+21)+(3x^2-4)=180

Solution for (2x^2+21)+(3x^2-4)=180 equation:

(2x^2+21)+(3x^2-4)=180

We move all terms to the left:

(2x^2+21)+(3x^2-4)-(180)=0

We get rid of parentheses

2x^2+3x^2+21-4-180=0

We add all the numbers together, and all the variables

5x^2-163=0

a = 5; b = 0; c = -163;
Δ = b2-4ac
Δ = 02-4·5·(-163)
Δ = 3260
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:

$\sqrt{\Delta}=\sqrt{3260}=\sqrt{4*815}=\sqrt{4}*\sqrt{815}=2\sqrt{815}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{815}}{2*5}=\frac{0-2\sqrt{815}}{10}
=-\frac{2\sqrt{815}}{10}
=-\frac{\sqrt{815}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{815}}{2*5}=\frac{0+2\sqrt{815}}{10}
=\frac{2\sqrt{815}}{10}
=\frac{\sqrt{815}}{5}
$