(2x^2-20)=(x^2+44)

Solution for (2x^2-20)=(x^2+44) equation:

(2x^2-20)=(x^2+44)

We move all terms to the left:

(2x^2-20)-((x^2+44))=0

We get rid of parentheses

2x^2-((x^2+44))-20=0


We calculate terms in parentheses: -((x^2+44)), so:

(x^2+44)

We get rid of parentheses

x^2+44

Back to the equation:

-(x^2+44)


We get rid of parentheses

2x^2-x^2-44-20=0

We add all the numbers together, and all the variables

x^2-64=0

a = 1; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·1·(-64)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*1}=\frac{16}{2}
=8
$