y^2+20y=10

Solution for y^2+20y=10 equation:

y^2+20y=10

We move all terms to the left:

y^2+20y-(10)=0

a = 1; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·1·(-10)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{110}}{2*1}=\frac{-20-2\sqrt{110}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{110}}{2*1}=\frac{-20+2\sqrt{110}}{2}
$