4q^2+3q=3q^2-14q+18

Solution for 4q^2+3q=3q^2-14q+18 equation:

4q^2+3q=3q^2-14q+18

We move all terms to the left:

4q^2+3q-(3q^2-14q+18)=0

We get rid of parentheses

4q^2-3q^2+3q+14q-18=0

We add all the numbers together, and all the variables

q^2+17q-18=0

a = 1; b = 17; c = -18;
Δ = b2-4ac
Δ = 172-4·1·(-18)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*1}=\frac{-36}{2}
=-18
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*1}=\frac{2}{2}
=1
$