w^2+11=9w^2-8

Solution for w^2+11=9w^2-8 equation:

w^2+11=9w^2-8

We move all terms to the left:

w^2+11-(9w^2-8)=0

We get rid of parentheses

w^2-9w^2+8+11=0

We add all the numbers together, and all the variables

-8w^2+19=0

a = -8; b = 0; c = +19;
Δ = b2-4ac
Δ = 02-4·(-8)·19
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{38}}{2*-8}=\frac{0-4\sqrt{38}}{-16}
=-\frac{4\sqrt{38}}{-16}
=-\frac{\sqrt{38}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{38}}{2*-8}=\frac{0+4\sqrt{38}}{-16}
=\frac{4\sqrt{38}}{-16}
=\frac{\sqrt{38}}{-4}
$