5(1+4h)h+2h=27

Solution for 5(1+4h)h+2h=27 equation:

5(1+4h)h+2h=27

We move all terms to the left:

5(1+4h)h+2h-(27)=0

We add all the numbers together, and all the variables

5(4h+1)h+2h-27=0

We add all the numbers together, and all the variables

2h+5(4h+1)h-27=0

We multiply parentheses

20h^2+2h+5h-27=0

We add all the numbers together, and all the variables

20h^2+7h-27=0

a = 20; b = 7; c = -27;
Δ = b2-4ac
Δ = 72-4·20·(-27)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-47}{2*20}=\frac{-54}{40}
=-1+7/20
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+47}{2*20}=\frac{40}{40}
=1
$