27=2x^2-5

Solution for 27=2x^2-5 equation:

27=2x^2-5

We move all terms to the left:

27-(2x^2-5)=0

We get rid of parentheses

-2x^2+5+27=0

We add all the numbers together, and all the variables

-2x^2+32=0

a = -2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-2)·32
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*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-2}=\frac{16}{-4}
=-4
$