q2+7q+12=0

Solution for q2+7q+12=0 equation:

q2+7q+12=0

We add all the numbers together, and all the variables

q^2+7q+12=0

a = 1; b = 7; c = +12;
Δ = b2-4ac
Δ = 72-4·1·12
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{-(7)-1}{2*1}=\frac{-8}{2}
=-4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*1}=\frac{-6}{2}
=-3
$