p^2-18p-79=0

Solution for p^2-18p-79=0 equation:

p^2-18p-79=0

a = 1; b = -18; c = -79;
Δ = b2-4ac
Δ = -182-4·1·(-79)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8\sqrt{10}}{2*1}=\frac{18-8\sqrt{10}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8\sqrt{10}}{2*1}=\frac{18+8\sqrt{10}}{2}
$