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

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

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

We move all terms to the left:

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

We multiply parentheses

4t-(4(2t-4)t)-12+8=0


We calculate terms in parentheses: -(4(2t-4)t), so:

4(2t-4)t

We multiply parentheses

8t^2-16t

Back to the equation:

-(8t^2-16t)


We add all the numbers together, and all the variables

4t-(8t^2-16t)-4=0

We get rid of parentheses

-8t^2+4t+16t-4=0

We add all the numbers together, and all the variables

-8t^2+20t-4=0

a = -8; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·(-8)·(-4)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{17}}{2*-8}=\frac{-20-4\sqrt{17}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{17}}{2*-8}=\frac{-20+4\sqrt{17}}{-16}
$