(3/2)h+1=2h+5

Solution for (3/2)h+1=2h+5 equation:

(3/2)h+1=2h+5

We move all terms to the left:

(3/2)h+1-(2h+5)=0


Domain of the equation: 2)h!=0

h!=0/1

h!=0

h∈R


We add all the numbers together, and all the variables

(+3/2)h-(2h+5)+1=0

We multiply parentheses

3h^2-(2h+5)+1=0

We get rid of parentheses

3h^2-2h-5+1=0

We add all the numbers together, and all the variables

3h^2-2h-4=0

a = 3; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·3·(-4)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*3}=\frac{2-2\sqrt{13}}{6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*3}=\frac{2+2\sqrt{13}}{6}
$