2x^2+7=41

Solution for 2x^2+7=41 equation:

2x^2+7=41

We move all terms to the left:

2x^2+7-(41)=0

We add all the numbers together, and all the variables

2x^2-34=0

a = 2; b = 0; c = -34;
Δ = b2-4ac
Δ = 02-4·2·(-34)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{17}}{2*2}=\frac{0-4\sqrt{17}}{4}
=-\frac{4\sqrt{17}}{4}
=-\sqrt{17}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{17}}{2*2}=\frac{0+4\sqrt{17}}{4}
=\frac{4\sqrt{17}}{4}
=\sqrt{17}
$