3/4h+3=-6h+18

Solution for 3/4h+3=-6h+18 equation:

3/4h+3=-6h+18

We move all terms to the left:

3/4h+3-(-6h+18)=0


Domain of the equation: 4h!=0

h!=0/4

h!=0

h∈R


We get rid of parentheses

3/4h+6h-18+3=0

We multiply all the terms by the denominator


6h*4h-18*4h+3*4h+3=0

Wy multiply elements

24h^2-72h+12h+3=0

We add all the numbers together, and all the variables

24h^2-60h+3=0

a = 24; b = -60; c = +3;
Δ = b2-4ac
Δ = -602-4·24·3
Δ = 3312
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:

$\sqrt{\Delta}=\sqrt{3312}=\sqrt{144*23}=\sqrt{144}*\sqrt{23}=12\sqrt{23}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-12\sqrt{23}}{2*24}=\frac{60-12\sqrt{23}}{48}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+12\sqrt{23}}{2*24}=\frac{60+12\sqrt{23}}{48}
$