T^3-5t^2-8t+3=0

Solution for T^3-5t^2-8t+3=0 equation:

^3-5T^2-8T+3=0

We add all the numbers together, and all the variables

-5T^2-8T=0

a = -5; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-5)·0
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{-(-8)-8}{2*-5}=\frac{0}{-10}
=0
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-5}=\frac{16}{-10}
=-1+3/5
$