5(1+4h)2h=27

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

5(1+4h)2h=27

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

40h^2+10h-27=0

a = 40; b = 10; c = -27;
Δ = b2-4ac
Δ = 102-4·40·(-27)
Δ = 4420
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:

$\sqrt{\Delta}=\sqrt{4420}=\sqrt{4*1105}=\sqrt{4}*\sqrt{1105}=2\sqrt{1105}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1105}}{2*40}=\frac{-10-2\sqrt{1105}}{80}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1105}}{2*40}=\frac{-10+2\sqrt{1105}}{80}
$