6y+10=(y^2)+3y

Solution for 6y+10=(y^2)+3y equation:

6y+10=(y^2)+3y

We move all terms to the left:

6y+10-((y^2)+3y)=0

We get rid of parentheses

-y^2+6y-3y+10=0

We add all the numbers together, and all the variables

-1y^2+3y+10=0

a = -1; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-1)·10
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*-1}=\frac{-10}{-2}
=+5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*-1}=\frac{4}{-2}
=-2
$