2x^2+11=40

Solution for 2x^2+11=40 equation:

2x^2+11=40

We move all terms to the left:

2x^2+11-(40)=0

We add all the numbers together, and all the variables

2x^2-29=0

a = 2; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·2·(-29)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{58}}{2*2}=\frac{0-2\sqrt{58}}{4}
=-\frac{2\sqrt{58}}{4}
=-\frac{\sqrt{58}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{58}}{2*2}=\frac{0+2\sqrt{58}}{4}
=\frac{2\sqrt{58}}{4}
=\frac{\sqrt{58}}{2}
$