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+3)=0


Domain of the equation: 5h!=0

h!=0/5

h!=0

h∈R



Domain of the equation: 5h-2h+3)!=0

h∈R


We add all the numbers together, and all the variables

2/5h-(-2h+12/5h+3)-7+=0

We add all the numbers together, and all the variables

2/5h-(-2h+12/5h+3)=0

We get rid of parentheses

2/5h+2h-12/5h-3=0

We multiply all the terms by the denominator


2h*5h-3*5h+2-12=0

We add all the numbers together, and all the variables

2h*5h-3*5h-10=0

Wy multiply elements

10h^2-15h-10=0

a = 10; b = -15; c = -10;
Δ = b2-4ac
Δ = -152-4·10·(-10)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-25}{2*10}=\frac{-10}{20}
=-1/2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*10}=\frac{40}{20}
=2
$