10=-1/3q^2+20

Solution for 10=-1/3q^2+20 equation:

10=-1/3q^2+20

We move all terms to the left:

10-(-1/3q^2+20)=0


Domain of the equation: 3q^2+20)!=0

q∈R


We get rid of parentheses

1/3q^2-20+10=0

We multiply all the terms by the denominator


-20*3q^2+10*3q^2+1=0

Wy multiply elements

-60q^2+30q^2+1=0

We add all the numbers together, and all the variables

-30q^2+1=0

a = -30; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-30)·1
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*-30}=\frac{0-2\sqrt{30}}{-60}
=-\frac{2\sqrt{30}}{-60}
=-\frac{\sqrt{30}}{-30}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*-30}=\frac{0+2\sqrt{30}}{-60}
=\frac{2\sqrt{30}}{-60}
=\frac{\sqrt{30}}{-30}
$