q^2+5q=14

Solution for q^2+5q=14 equation:

q^2+5q=14

We move all terms to the left:

q^2+5q-(14)=0

a = 1; b = 5; c = -14;
Δ = b2-4ac
Δ = 52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*1}=\frac{-14}{2}
=-7
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*1}=\frac{4}{2}
=2
$