p^2-14p=5

Solution for p^2-14p=5 equation:

p^2-14p=5

We move all terms to the left:

p^2-14p-(5)=0

a = 1; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·1·(-5)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{6}}{2*1}=\frac{14-6\sqrt{6}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{6}}{2*1}=\frac{14+6\sqrt{6}}{2}
$