(4/5)h=2

Solution for (4/5)h=2 equation:

(4/5)h=2

We move all terms to the left:

(4/5)h-(2)=0


Domain of the equation: 5)h!=0

h!=0/1

h!=0

h∈R


We add all the numbers together, and all the variables

(+4/5)h-2=0

We multiply parentheses

4h^2-2=0

a = 4; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·4·(-2)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*4}=\frac{0-4\sqrt{2}}{8}
=-\frac{4\sqrt{2}}{8}
=-\frac{\sqrt{2}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*4}=\frac{0+4\sqrt{2}}{8}
=\frac{4\sqrt{2}}{8}
=\frac{\sqrt{2}}{2}
$