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

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

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

We move all terms to the left:

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


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

q^2!=0/3

q^2!=√0

q!=0

q∈R



Domain of the equation: 3q^2+10)!=0

q∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-10*3q^2+2=0

Wy multiply elements

-30q^2+2=0

a = -30; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-30)·2
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-30}=\frac{0-4\sqrt{15}}{-60}
=-\frac{4\sqrt{15}}{-60}
=-\frac{\sqrt{15}}{-15}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-30}=\frac{0+4\sqrt{15}}{-60}
=\frac{4\sqrt{15}}{-60}
=\frac{\sqrt{15}}{-15}
$