Y=1/2x^-3x+4

Solution for Y=1/2x^-3x+4 equation:

=1/2Y^-3Y+4

We move all terms to the left:

-(1/2Y^-3Y+4)=0


Domain of the equation: 2Y^-3Y+4)!=0

Y∈R


We get rid of parentheses

-1/2Y^+3Y-4=0

We multiply all the terms by the denominator


3Y*2Y^-4*2Y^-1=0

Wy multiply elements

6Y^2-8Y-1=0

a = 6; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·6·(-1)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{22}}{2*6}=\frac{8-2\sqrt{22}}{12}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{22}}{2*6}=\frac{8+2\sqrt{22}}{12}
$