7=10/5h-2h+3

Solution for 7=10/5h-2h+3 equation:

7=10/5h-2h+3

We move all terms to the left:

7-(10/5h-2h+3)=0


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

h∈R


We add all the numbers together, and all the variables

-(-2h+10/5h+3)+7=0

We get rid of parentheses

2h-10/5h-3+7=0

We multiply all the terms by the denominator


2h*5h-3*5h+7*5h-10=0

Wy multiply elements

10h^2-15h+35h-10=0

We add all the numbers together, and all the variables

10h^2+20h-10=0

a = 10; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·10·(-10)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*10}=\frac{-20-20\sqrt{2}}{20}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*10}=\frac{-20+20\sqrt{2}}{20}
$