p^2-34p+24=0

Solution for p^2-34p+24=0 equation:

p^2-34p+24=0

a = 1; b = -34; c = +24;
Δ = b2-4ac
Δ = -342-4·1·24
Δ = 1060
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:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{265}}{2*1}=\frac{34-2\sqrt{265}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{265}}{2*1}=\frac{34+2\sqrt{265}}{2}
$