1/3q+8=q-10

Solution for 1/3q+8=q-10 equation:

1/3q+8=q-10

We move all terms to the left:

1/3q+8-(q-10)=0


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R


We get rid of parentheses

1/3q-q+10+8=0

We multiply all the terms by the denominator


-q*3q+10*3q+8*3q+1=0

Wy multiply elements

-3q^2+30q+24q+1=0

We add all the numbers together, and all the variables

-3q^2+54q+1=0

a = -3; b = 54; c = +1;
Δ = b2-4ac
Δ = 542-4·(-3)·1
Δ = 2928
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:

$\sqrt{\Delta}=\sqrt{2928}=\sqrt{16*183}=\sqrt{16}*\sqrt{183}=4\sqrt{183}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-4\sqrt{183}}{2*-3}=\frac{-54-4\sqrt{183}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+4\sqrt{183}}{2*-3}=\frac{-54+4\sqrt{183}}{-6}
$