2m^2-5=27

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

2m^2-5=27

We move all terms to the left:

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

We add all the numbers together, and all the variables

2m^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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4}
=4
$