p2+13p+12=0

Solution for p2+13p+12=0 equation:

p2+13p+12=0

We add all the numbers together, and all the variables

p^2+13p+12=0

a = 1; b = 13; c = +12;
Δ = b2-4ac
Δ = 132-4·1·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*1}=\frac{-24}{2}
=-12
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*1}=\frac{-2}{2}
=-1
$