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+9y+10=0

a = -1; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·(-1)·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*-1}=\frac{-20}{-2}
=+10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*-1}=\frac{2}{-2}
=-1
$