3/5h-7=17/5h-2h

Solution for 3/5h-7=17/5h-2h equation:

3/5h-7=17/5h-2h

We move all terms to the left:

3/5h-7-(17/5h-2h)=0


Domain of the equation: 5h!=0

h!=0/5

h!=0

h∈R



Domain of the equation: 5h-2h)!=0

h∈R


We add all the numbers together, and all the variables

3/5h-(-2h+17/5h)-7=0

We get rid of parentheses

3/5h+2h-17/5h-7=0

We multiply all the terms by the denominator


2h*5h-7*5h+3-17=0

We add all the numbers together, and all the variables

2h*5h-7*5h-14=0

Wy multiply elements

10h^2-35h-14=0

a = 10; b = -35; c = -14;
Δ = b2-4ac
Δ = -352-4·10·(-14)
Δ = 1785
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{1785}}{2*10}=\frac{35-\sqrt{1785}}{20}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1785}}{2*10}=\frac{35+\sqrt{1785}}{20}
$