5h-2;3/2h+4)=10

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

5h-23/2h+4)=10

We move all terms to the left:

5h-23/2h+4)-(10)=0


Domain of the equation: 2h!=0

h!=0/2

h!=0

h∈R


We add all the numbers together, and all the variables

5h-23/2h=0

We multiply all the terms by the denominator


5h*2h-23=0

Wy multiply elements

10h^2-23=0

a = 10; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·10·(-23)
Δ = 920
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:

$\sqrt{\Delta}=\sqrt{920}=\sqrt{4*230}=\sqrt{4}*\sqrt{230}=2\sqrt{230}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{230}}{2*10}=\frac{0-2\sqrt{230}}{20}
=-\frac{2\sqrt{230}}{20}
=-\frac{\sqrt{230}}{10}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{230}}{2*10}=\frac{0+2\sqrt{230}}{20}
=\frac{2\sqrt{230}}{20}
=\frac{\sqrt{230}}{10}
$