5=-1/3q^2+10

Solution for 5=-1/3q^2+10 equation:

5=-1/3q^2+10

We move all terms to the left:

5-(-1/3q^2+10)=0


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

q∈R


We get rid of parentheses

1/3q^2-10+5=0

We multiply all the terms by the denominator


-10*3q^2+5*3q^2+1=0

Wy multiply elements

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

We add all the numbers together, and all the variables

-15q^2+1=0

a = -15; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-15)·1
Δ = 60
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:

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