5y^2+6y+3=6y-4y^2+4

Solution for 5y^2+6y+3=6y-4y^2+4 equation:

5y^2+6y+3=6y-4y^2+4

We move all terms to the left:

5y^2+6y+3-(6y-4y^2+4)=0

We get rid of parentheses

5y^2+4y^2-6y+6y-4+3=0

We add all the numbers together, and all the variables

9y^2-1=0

a = 9; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·9·(-1)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*9}=\frac{-6}{18}
=-1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*9}=\frac{6}{18}
=1/3
$