4q^2-20q+14=22+11q

Solution for 4q^2-20q+14=22+11q equation:

4q^2-20q+14=22+11q

We move all terms to the left:

4q^2-20q+14-(22+11q)=0

We add all the numbers together, and all the variables

4q^2-20q-(11q+22)+14=0

We get rid of parentheses

4q^2-20q-11q-22+14=0

We add all the numbers together, and all the variables

4q^2-31q-8=0

a = 4; b = -31; c = -8;
Δ = b2-4ac
Δ = -312-4·4·(-8)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-33}{2*4}=\frac{-2}{8}
=-1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+33}{2*4}=\frac{64}{8}
=8
$