p2+-7p+12=0

Solution for p2+-7p+12=0 equation:

p2+-7p+12=0

We add all the numbers together, and all the variables

p^2-7p=0

a = 1; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·1·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*1}=\frac{0}{2}
=0
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*1}=\frac{14}{2}
=7
$