3h-2(4h-5)=10/5h

Solution for 3h-2(4h-5)=10/5h equation:

3h-2(4h-5)=10/5h

We move all terms to the left:

3h-2(4h-5)-(10/5h)=0


Domain of the equation: 5h)!=0

h!=0/1

h!=0

h∈R


We add all the numbers together, and all the variables

3h-2(4h-5)-(+10/5h)=0

We multiply parentheses

3h-8h-(+10/5h)+10=0

We get rid of parentheses

3h-8h-10/5h+10=0

We multiply all the terms by the denominator


3h*5h-8h*5h+10*5h-10=0

Wy multiply elements

15h^2-40h^2+50h-10=0

We add all the numbers together, and all the variables

-25h^2+50h-10=0

a = -25; b = 50; c = -10;
Δ = b2-4ac
Δ = 502-4·(-25)·(-10)
Δ = 1500
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:

$\sqrt{\Delta}=\sqrt{1500}=\sqrt{100*15}=\sqrt{100}*\sqrt{15}=10\sqrt{15}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{15}}{2*-25}=\frac{-50-10\sqrt{15}}{-50}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{15}}{2*-25}=\frac{-50+10\sqrt{15}}{-50}
$