(2a^2-54)=(a^2+27)

Solution for (2a^2-54)=(a^2+27) equation:

(2a^2-54)=(a^2+27)

We move all terms to the left:

(2a^2-54)-((a^2+27))=0

We get rid of parentheses

2a^2-((a^2+27))-54=0


We calculate terms in parentheses: -((a^2+27)), so:

(a^2+27)

We get rid of parentheses

a^2+27

Back to the equation:

-(a^2+27)


We get rid of parentheses

2a^2-a^2-27-54=0

We add all the numbers together, and all the variables

a^2-81=0

a = 1; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·1·(-81)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*1}=\frac{-18}{2}
=-9
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*1}=\frac{18}{2}
=9
$