p^2-12p+14=-4p-1

Solution for p^2-12p+14=-4p-1 equation:

p^2-12p+14=-4p-1

We move all terms to the left:

p^2-12p+14-(-4p-1)=0

We get rid of parentheses

p^2-12p+4p+1+14=0

We add all the numbers together, and all the variables

p^2-8p+15=0

a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2}
=3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2}
=5
$