13(y^2)-18y+3=9(y^2)-10y

Solution for 13(y^2)-18y+3=9(y^2)-10y equation:

13(y^2)-18y+3=9(y^2)-10y

We move all terms to the left:

13(y^2)-18y+3-(9(y^2)-10y)=0

We get rid of parentheses

13y^2-9y^2-18y+10y+3=0

We add all the numbers together, and all the variables

4y^2-8y+3=0

a = 4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·4·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*4}=\frac{4}{8}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8}
=1+1/2
$