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

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

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

We move all terms to the left:

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


Domain of the equation: 5h!=0

h!=0/5

h!=0

h∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-50h^2-15h-35h+2=0

We add all the numbers together, and all the variables

-50h^2-50h+2=0

a = -50; b = -50; c = +2;
Δ = b2-4ac
Δ = -502-4·(-50)·2
Δ = 2900
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:

$\sqrt{\Delta}=\sqrt{2900}=\sqrt{100*29}=\sqrt{100}*\sqrt{29}=10\sqrt{29}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{29}}{2*-50}=\frac{50-10\sqrt{29}}{-100}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{29}}{2*-50}=\frac{50+10\sqrt{29}}{-100}
$