4h^2-40h+99=0

Solution for 4h^2-40h+99=0 equation:

4h^2-40h+99=0

a = 4; b = -40; c = +99;
Δ = b2-4ac
Δ = -402-4·4·99
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4}{2*4}=\frac{36}{8}
=4+1/2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4}{2*4}=\frac{44}{8}
=5+1/2
$