(2y^2+4)=(y^2+40)

Solution for (2y^2+4)=(y^2+40) equation:

(2y^2+4)=(y^2+40)

We move all terms to the left:

(2y^2+4)-((y^2+40))=0

We get rid of parentheses

2y^2-((y^2+40))+4=0


We calculate terms in parentheses: -((y^2+40)), so:

(y^2+40)

We get rid of parentheses

y^2+40

Back to the equation:

-(y^2+40)


We get rid of parentheses

2y^2-y^2-40+4=0

We add all the numbers together, and all the variables

y^2-36=0

a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*1}=\frac{-12}{2}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*1}=\frac{12}{2}
=6
$