p^2+6p=40

Solution for p^2+6p=40 equation:

p^2+6p=40

We move all terms to the left:

p^2+6p-(40)=0

a = 1; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*1}=\frac{-20}{2}
=-10
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*1}=\frac{8}{2}
=4
$