2(2t+4)=(3/4)(24-8t)

Solution for 2(2t+4)=(3/4)(24-8t) equation:

2(2t+4)=(3/4)(24-8t)

We move all terms to the left:

2(2t+4)-((3/4)(24-8t))=0


Domain of the equation: 4)(24-8t))!=0

t∈R


We add all the numbers together, and all the variables

2(2t+4)-((+3/4)(-8t+24))=0

We multiply parentheses

4t-((+3/4)(-8t+24))+8=0

We multiply parentheses ..

-((-24t^2+3/4*24))+4t+8=0

We multiply all the terms by the denominator


-((-24t^2+3+4t*4*24))+8*4*24))=0


We calculate terms in parentheses: -((-24t^2+3+4t*4*24)), so:

(-24t^2+3+4t*4*24)

We get rid of parentheses

-24t^2+4t*4*24+3

Wy multiply elements

-24t^2+384t*2+3

Wy multiply elements

-24t^2+768t+3

Back to the equation:

-(-24t^2+768t+3)


We add all the numbers together, and all the variables

-(-24t^2+768t+3)=0

We get rid of parentheses

24t^2-768t-3=0

a = 24; b = -768; c = -3;
Δ = b2-4ac
Δ = -7682-4·24·(-3)
Δ = 590112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{590112}=\sqrt{144*4098}=\sqrt{144}*\sqrt{4098}=12\sqrt{4098}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-768)-12\sqrt{4098}}{2*24}=\frac{768-12\sqrt{4098}}{48}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-768)+12\sqrt{4098}}{2*24}=\frac{768+12\sqrt{4098}}{48}
$