3t-2t(t-1)=5t-4(2+t)

Solution for 3t-2t(t-1)=5t-4(2+t) equation:

3t-2t(t-1)=5t-4(2+t)

We move all terms to the left:

3t-2t(t-1)-(5t-4(2+t))=0

We add all the numbers together, and all the variables

3t-2t(t-1)-(5t-4(t+2))=0

We multiply parentheses

-2t^2+3t+2t-(5t-4(t+2))=0


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

5t-4(t+2)

We multiply parentheses

5t-4t-8

We add all the numbers together, and all the variables

t-8

Back to the equation:

-(t-8)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-2t^2+5t-t+8=0

We add all the numbers together, and all the variables

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

a = -2; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-2)·8
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{5}}{2*-2}=\frac{-4-4\sqrt{5}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{5}}{2*-2}=\frac{-4+4\sqrt{5}}{-4}
$