8q^2-3q=7q-6q+18

Solution for 8q^2-3q=7q-6q+18 equation:

8q^2-3q=7q-6q+18

We move all terms to the left:

8q^2-3q-(7q-6q+18)=0

We add all the numbers together, and all the variables

8q^2-3q-(q+18)=0

We get rid of parentheses

8q^2-3q-q-18=0

We add all the numbers together, and all the variables

8q^2-4q-18=0

a = 8; b = -4; c = -18;
Δ = b2-4ac
Δ = -42-4·8·(-18)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{37}}{2*8}=\frac{4-4\sqrt{37}}{16}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{37}}{2*8}=\frac{4+4\sqrt{37}}{16}
$