p^2-14p+52=7

Solution for p^2-14p+52=7 equation:

p^2-14p+52=7

We move all terms to the left:

p^2-14p+52-(7)=0

We add all the numbers together, and all the variables

p^2-14p+45=0

a = 1; b = -14; c = +45;
Δ = b2-4ac
Δ = -142-4·1·45
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*1}=\frac{10}{2}
=5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*1}=\frac{18}{2}
=9
$