3y^2-8y+13=2y^2

Solution for 3y^2-8y+13=2y^2 equation:

3y^2-8y+13=2y^2

We move all terms to the left:

3y^2-8y+13-(2y^2)=0

determiningTheFunctionDomain
3y^2-2y^2-8y+13=0

We add all the numbers together, and all the variables

y^2-8y+13=0

a = 1; b = -8; c = +13;
Δ = b2-4ac
Δ = -82-4·1·13
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{3}}{2*1}=\frac{8-2\sqrt{3}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{3}}{2*1}=\frac{8+2\sqrt{3}}{2}
$