p2+15p+56=0

Solution for p2+15p+56=0 equation:

p2+15p+56=0

We add all the numbers together, and all the variables

p^2+15p+56=0

a = 1; b = 15; c = +56;
Δ = b2-4ac
Δ = 152-4·1·56
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*1}=\frac{-16}{2}
=-8
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*1}=\frac{-14}{2}
=-7
$