9y^2-3y+4=-12+4y+6+13y

Solution for 9y^2-3y+4=-12+4y+6+13y equation:

9y^2-3y+4=-12+4y+6+13y

We move all terms to the left:

9y^2-3y+4-(-12+4y+6+13y)=0

We add all the numbers together, and all the variables

9y^2-3y-(17y-6)+4=0

We get rid of parentheses

9y^2-3y-17y+6+4=0

We add all the numbers together, and all the variables

9y^2-20y+10=0

a = 9; b = -20; c = +10;
Δ = b2-4ac
Δ = -202-4·9·10
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{10}}{2*9}=\frac{20-2\sqrt{10}}{18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{10}}{2*9}=\frac{20+2\sqrt{10}}{18}
$