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

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

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

We move all terms to the left:

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


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

t∈R


We add all the numbers together, and all the variables

2(2t+4)-(3/416t)=0

We multiply parentheses

4t-(3/416t)+8=0

We get rid of parentheses

4t-3/416t+8=0

We multiply all the terms by the denominator


4t*416t+8*416t-3=0

Wy multiply elements

1664t^2+3328t-3=0

a = 1664; b = 3328; c = -3;
Δ = b2-4ac
Δ = 33282-4·1664·(-3)
Δ = 11095552
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:

$\sqrt{\Delta}=\sqrt{11095552}=\sqrt{256*43342}=\sqrt{256}*\sqrt{43342}=16\sqrt{43342}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3328)-16\sqrt{43342}}{2*1664}=\frac{-3328-16\sqrt{43342}}{3328}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3328)+16\sqrt{43342}}{2*1664}=\frac{-3328+16\sqrt{43342}}{3328}
$