T=(3t-5)(t+1)

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

=(3T-5)(T+1)

We move all terms to the left:

-((3T-5)(T+1))=0

We multiply parentheses ..

-((+3T^2+3T-5T-5))=0


We calculate terms in parentheses: -((+3T^2+3T-5T-5)), so:

(+3T^2+3T-5T-5)

We get rid of parentheses

3T^2+3T-5T-5

We add all the numbers together, and all the variables

3T^2-2T-5

Back to the equation:

-(3T^2-2T-5)


We get rid of parentheses

-3T^2+2T+5=0

a = -3; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-3)·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-3}=\frac{6}{-6}
=-1
$