p^2-34p+225=0

Solution for p^2-34p+225=0 equation:

p^2-34p+225=0

a = 1; b = -34; c = +225;
Δ = b2-4ac
Δ = -342-4·1·225
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-16}{2*1}=\frac{18}{2}
=9
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+16}{2*1}=\frac{50}{2}
=25
$