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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

4(2t-4)

We multiply parentheses

8t-16

Back to the equation:

-(8t-16)


We get rid of parentheses

4t^2-12t-8t+16+8=0

We add all the numbers together, and all the variables

4t^2-20t+24=0

a = 4; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*4}=\frac{16}{8}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*4}=\frac{24}{8}
=3
$