q=3/2q-9

Solution for q=3/2q-9 equation:

q=3/2q-9

We move all terms to the left:

q-(3/2q-9)=0


Domain of the equation: 2q-9)!=0

q∈R


We get rid of parentheses

q-3/2q+9=0

We multiply all the terms by the denominator


q*2q+9*2q-3=0

Wy multiply elements

2q^2+18q-3=0

a = 2; b = 18; c = -3;
Δ = b2-4ac
Δ = 182-4·2·(-3)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{87}}{2*2}=\frac{-18-2\sqrt{87}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{87}}{2*2}=\frac{-18+2\sqrt{87}}{4}
$