(5y^2-3)+2=2y^2

Solution for (5y^2-3)+2=2y^2 equation:

(5y^2-3)+2=2y^2

We move all terms to the left:

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

determiningTheFunctionDomain
-2y^2+(5y^2-3)+2=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

3y^2-1=0

a = 3; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·3·(-1)
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{-(0)-2\sqrt{3}}{2*3}=\frac{0-2\sqrt{3}}{6}
=-\frac{2\sqrt{3}}{6}
=-\frac{\sqrt{3}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*3}=\frac{0+2\sqrt{3}}{6}
=\frac{2\sqrt{3}}{6}
=\frac{\sqrt{3}}{3}
$