y(y+4)+3(y-3)=10

Solution for y(y+4)+3(y-3)=10 equation:

y(y+4)+3(y-3)=10

We move all terms to the left:

y(y+4)+3(y-3)-(10)=0

We multiply parentheses

y^2+4y+3y-9-10=0

We add all the numbers together, and all the variables

y^2+7y-19=0

a = 1; b = 7; c = -19;
Δ = b2-4ac
Δ = 72-4·1·(-19)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5\sqrt{5}}{2*1}=\frac{-7-5\sqrt{5}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5\sqrt{5}}{2*1}=\frac{-7+5\sqrt{5}}{2}
$