-3/4h=-h-3-12

Solution for -3/4h=-h-3-12 equation:

-3/4h=-h-3-12

We move all terms to the left:

-3/4h-(-h-3-12)=0


Domain of the equation: 4h!=0

h!=0/4

h!=0

h∈R


We add all the numbers together, and all the variables

-3/4h-(-1h-15)=0

We get rid of parentheses

-3/4h+1h+15=0

We multiply all the terms by the denominator


1h*4h+15*4h-3=0

Wy multiply elements

4h^2+60h-3=0

a = 4; b = 60; c = -3;
Δ = b2-4ac
Δ = 602-4·4·(-3)
Δ = 3648
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:

$\sqrt{\Delta}=\sqrt{3648}=\sqrt{64*57}=\sqrt{64}*\sqrt{57}=8\sqrt{57}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-8\sqrt{57}}{2*4}=\frac{-60-8\sqrt{57}}{8}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+8\sqrt{57}}{2*4}=\frac{-60+8\sqrt{57}}{8}
$