(y^2)+10+25=233

Solution for (y^2)+10+25=233 equation:

(y^2)+10+25=233

We move all terms to the left:

(y^2)+10+25-(233)=0

We add all the numbers together, and all the variables

y^2-198=0

a = 1; b = 0; c = -198;
Δ = b2-4ac
Δ = 02-4·1·(-198)
Δ = 792
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{22}}{2*1}=\frac{0-6\sqrt{22}}{2}
=-\frac{6\sqrt{22}}{2}
=-3\sqrt{22}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{22}}{2*1}=\frac{0+6\sqrt{22}}{2}
=\frac{6\sqrt{22}}{2}
=3\sqrt{22}
$