(2x^2+7)=13

Solution for (2x^2+7)=13 equation:

(2x^2+7)=13

We move all terms to the left:

(2x^2+7)-(13)=0

We get rid of parentheses

2x^2+7-13=0

We add all the numbers together, and all the variables

2x^2-6=0

a = 2; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·2·(-6)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*2}=\frac{0-4\sqrt{3}}{4}
=-\frac{4\sqrt{3}}{4}
=-\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*2}=\frac{0+4\sqrt{3}}{4}
=\frac{4\sqrt{3}}{4}
=\sqrt{3}
$