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

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

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

We move all terms to the left:

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


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

h∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


5h*5h+2h*5h-3*5h-7*5h-12=0

Wy multiply elements

25h^2+10h^2-15h-35h-12=0

We add all the numbers together, and all the variables

35h^2-50h-12=0

a = 35; b = -50; c = -12;
Δ = b2-4ac
Δ = -502-4·35·(-12)
Δ = 4180
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:

$\sqrt{\Delta}=\sqrt{4180}=\sqrt{4*1045}=\sqrt{4}*\sqrt{1045}=2\sqrt{1045}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{1045}}{2*35}=\frac{50-2\sqrt{1045}}{70}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{1045}}{2*35}=\frac{50+2\sqrt{1045}}{70}
$