p2+p-42=0

Solution for p2+p-42=0 equation:

p2+p-42=0

We add all the numbers together, and all the variables

p^2+p-42=0

a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2}
=-7
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2}
=6
$