-9v^2+18=11

Solution for -9v^2+18=11 equation:

-9v^2+18=11

We move all terms to the left:

-9v^2+18-(11)=0

We add all the numbers together, and all the variables

-9v^2+7=0

a = -9; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-9)·7
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*-9}=\frac{0-6\sqrt{7}}{-18}
=-\frac{6\sqrt{7}}{-18}
=-\frac{\sqrt{7}}{-3}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*-9}=\frac{0+6\sqrt{7}}{-18}
=\frac{6\sqrt{7}}{-18}
=\frac{\sqrt{7}}{-3}
$