3(h^2-4)=5h(h-1)-9h

Solution for 3(h^2-4)=5h(h-1)-9h equation:

3(h^2-4)=5h(h-1)-9h

We move all terms to the left:

3(h^2-4)-(5h(h-1)-9h)=0

We multiply parentheses

3h^2-(5h(h-1)-9h)-12=0


We calculate terms in parentheses: -(5h(h-1)-9h), so:

5h(h-1)-9h

We add all the numbers together, and all the variables

-9h+5h(h-1)

We multiply parentheses

5h^2-9h-5h

We add all the numbers together, and all the variables

5h^2-14h

Back to the equation:

-(5h^2-14h)


We get rid of parentheses

3h^2-5h^2+14h-12=0

We add all the numbers together, and all the variables

-2h^2+14h-12=0

a = -2; b = 14; c = -12;
Δ = b2-4ac
Δ = 142-4·(-2)·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*-2}=\frac{-24}{-4}
=+6
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*-2}=\frac{-4}{-4}
=1
$