2x^2+5=36

Solution for 2x^2+5=36 equation:

2x^2+5=36

We move all terms to the left:

2x^2+5-(36)=0

We add all the numbers together, and all the variables

2x^2-31=0

a = 2; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·2·(-31)
Δ = 248
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*2}=\frac{0-2\sqrt{62}}{4}
=-\frac{2\sqrt{62}}{4}
=-\frac{\sqrt{62}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*2}=\frac{0+2\sqrt{62}}{4}
=\frac{2\sqrt{62}}{4}
=\frac{\sqrt{62}}{2}
$