p^2-28p-196=0

Solution for p^2-28p-196=0 equation:

p^2-28p-196=0

a = 1; b = -28; c = -196;
Δ = b2-4ac
Δ = -282-4·1·(-196)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28\sqrt{2}}{2*1}=\frac{28-28\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28\sqrt{2}}{2*1}=\frac{28+28\sqrt{2}}{2}
$