2q^2-q+4=6q^2-2q+4

Solution for 2q^2-q+4=6q^2-2q+4 equation:

2q^2-q+4=6q^2-2q+4

We move all terms to the left:

2q^2-q+4-(6q^2-2q+4)=0

We add all the numbers together, and all the variables

2q^2-1q-(6q^2-2q+4)+4=0

We get rid of parentheses

2q^2-6q^2-1q+2q-4+4=0

We add all the numbers together, and all the variables

-4q^2+q=0

a = -4; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-4}=\frac{-2}{-8}
=1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-4}=\frac{0}{-8}
=0
$