4(t-2)+4=4t(2t-5)

Solution for 4(t-2)+4=4t(2t-5) equation:

4(t-2)+4=4t(2t-5)

We move all terms to the left:

4(t-2)+4-(4t(2t-5))=0

We multiply parentheses

4t-(4t(2t-5))-8+4=0


We calculate terms in parentheses: -(4t(2t-5)), so:

4t(2t-5)

We multiply parentheses

8t^2-20t

Back to the equation:

-(8t^2-20t)


We add all the numbers together, and all the variables

4t-(8t^2-20t)-4=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-8t^2+24t-4=0

a = -8; b = 24; c = -4;
Δ = b2-4ac
Δ = 242-4·(-8)·(-4)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{7}}{2*-8}=\frac{-24-8\sqrt{7}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{7}}{2*-8}=\frac{-24+8\sqrt{7}}{-16}
$