2x^2+13=41

Solution for 2x^2+13=41 equation:

2x^2+13=41

We move all terms to the left:

2x^2+13-(41)=0

We add all the numbers together, and all the variables

2x^2-28=0

a = 2; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·2·(-28)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*2}=\frac{0-4\sqrt{14}}{4}
=-\frac{4\sqrt{14}}{4}
=-\sqrt{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*2}=\frac{0+4\sqrt{14}}{4}
=\frac{4\sqrt{14}}{4}
=\sqrt{14}
$