h^2-14h+20=0

Solution for h^2-14h+20=0 equation:

h^2-14h+20=0

a = 1; b = -14; c = +20;
Δ = b2-4ac
Δ = -142-4·1·20
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{29}}{2*1}=\frac{14-2\sqrt{29}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{29}}{2*1}=\frac{14+2\sqrt{29}}{2}
$