2/5h-7=12/5h-2h=3

Solution for 2/5h-7=12/5h-2h=3 equation:

2/5h-7=12/5h-2h=3

We move all terms to the left:

2/5h-7-(12/5h-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

2/5h-(-2h+12/5h)-7=0

We get rid of parentheses

2/5h+2h-12/5h-7=0

We multiply all the terms by the denominator


2h*5h-7*5h+2-12=0

We add all the numbers together, and all the variables

2h*5h-7*5h-10=0

Wy multiply elements

10h^2-35h-10=0

a = 10; b = -35; c = -10;
Δ = b2-4ac
Δ = -352-4·10·(-10)
Δ = 1625
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:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{65}}{2*10}=\frac{35-5\sqrt{65}}{20}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{65}}{2*10}=\frac{35+5\sqrt{65}}{20}
$