15=-1/3q^2+30

Solution for 15=-1/3q^2+30 equation:

15=-1/3q^2+30

We move all terms to the left:

15-(-1/3q^2+30)=0


Domain of the equation: 3q^2+30)!=0

q∈R


We get rid of parentheses

1/3q^2-30+15=0

We multiply all the terms by the denominator


-30*3q^2+15*3q^2+1=0

Wy multiply elements

-90q^2+45q^2+1=0

We add all the numbers together, and all the variables

-45q^2+1=0

a = -45; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-45)·1
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-45}=\frac{0-6\sqrt{5}}{-90}
=-\frac{6\sqrt{5}}{-90}
=-\frac{\sqrt{5}}{-15}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-45}=\frac{0+6\sqrt{5}}{-90}
=\frac{6\sqrt{5}}{-90}
=\frac{\sqrt{5}}{-15}
$