0=9q^2-15+4

Solution for 0=9q^2-15+4 equation:

0=9q^2-15+4

We move all terms to the left:

0-(9q^2-15+4)=0

We add all the numbers together, and all the variables

-(9q^2-15+4)=0

We get rid of parentheses

-9q^2+15-4=0

We add all the numbers together, and all the variables

-9q^2+11=0

a = -9; b = 0; c = +11;
Δ = b2-4ac
Δ = 02-4·(-9)·11
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{11}}{2*-9}=\frac{0-6\sqrt{11}}{-18}
=-\frac{6\sqrt{11}}{-18}
=-\frac{\sqrt{11}}{-3}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{11}}{2*-9}=\frac{0+6\sqrt{11}}{-18}
=\frac{6\sqrt{11}}{-18}
=\frac{\sqrt{11}}{-3}
$