Y^3-4y2+6y-5=0

Solution for Y^3-4y2+6y-5=0 equation:

^3-4Y^2+6Y-5=0

We add all the numbers together, and all the variables

-4Y^2+6Y=0

a = -4; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-4)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-4}=\frac{0}{-8}
=0
$