y(3+y)=130

Solution for y(3+y)=130 equation:

y(3+y)=130

We move all terms to the left:

y(3+y)-(130)=0

We add all the numbers together, and all the variables

y(y+3)-130=0

We multiply parentheses

y^2+3y-130=0

a = 1; b = 3; c = -130;
Δ = b2-4ac
Δ = 32-4·1·(-130)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*1}=\frac{-26}{2}
=-13
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*1}=\frac{20}{2}
=10
$