(Y^3+1)-((y^2+1)+(3y-7))=0

Solution for (Y^3+1)-((y^2+1)+(3y-7))=0 equation:

(^3+1)-((Y^2+1)+(3Y-7))=0

We add all the numbers together, and all the variables

-((Y^2+1)+(3Y-7))=0


We calculate terms in parentheses: -((Y^2+1)+(3Y-7)), so:

(Y^2+1)+(3Y-7)

We get rid of parentheses

Y^2+3Y+1-7

We add all the numbers together, and all the variables

Y^2+3Y-6

Back to the equation:

-(Y^2+3Y-6)


We get rid of parentheses

-Y^2-3Y+6=0

We add all the numbers together, and all the variables

-1Y^2-3Y+6=0

a = -1; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-1)·6
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*-1}=\frac{3-\sqrt{33}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*-1}=\frac{3+\sqrt{33}}{-2}
$