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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

y^2+10y-208=0

a = 1; b = 10; c = -208;
Δ = b2-4ac
Δ = 102-4·1·(-208)
Δ = 932
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:

$\sqrt{\Delta}=\sqrt{932}=\sqrt{4*233}=\sqrt{4}*\sqrt{233}=2\sqrt{233}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{233}}{2*1}=\frac{-10-2\sqrt{233}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{233}}{2*1}=\frac{-10+2\sqrt{233}}{2}
$