q2+16q+15=0

Solution for q2+16q+15=0 equation:

q2+16q+15=0

We add all the numbers together, and all the variables

q^2+16q+15=0

a = 1; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·1·15
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*1}=\frac{-30}{2}
=-15
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*1}=\frac{-2}{2}
=-1
$