2q2+10q+12=q2

Solution for 2q2+10q+12=q2 equation:

2q^2+10q+12=q2

We move all terms to the left:

2q^2+10q+12-(q2)=0

We add all the numbers together, and all the variables

q^2+10q+12=0

a = 1; b = 10; c = +12;
Δ = b2-4ac
Δ = 102-4·1·12
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{13}}{2*1}=\frac{-10-2\sqrt{13}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{13}}{2*1}=\frac{-10+2\sqrt{13}}{2}
$