2w^2-11-40=0

Solution for 2w^2-11-40=0 equation:

2w^2-11-40=0

We add all the numbers together, and all the variables

2w^2-51=0

a = 2; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·2·(-51)
Δ = 408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{102}}{2*2}=\frac{0-2\sqrt{102}}{4}
=-\frac{2\sqrt{102}}{4}
=-\frac{\sqrt{102}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{102}}{2*2}=\frac{0+2\sqrt{102}}{4}
=\frac{2\sqrt{102}}{4}
=\frac{\sqrt{102}}{2}
$