2/z-5=0.5z2+7

Solution for 2/z-5=0.5z2+7 equation:

2/z-5=0.5z^2+7

We move all terms to the left:

2/z-5-(0.5z^2+7)=0


Domain of the equation: z!=0

z∈R


We get rid of parentheses

-0.5z^2+2/z-7-5=0

We multiply all the terms by the denominator


-(0.5z^2)*z-7*z-5*z+2=0

We add all the numbers together, and all the variables

-12z-(0.5z^2)*z+2=0

We multiply parentheses

-0z^2-12z+2=0

We add all the numbers together, and all the variables

-1z^2-12z+2=0

a = -1; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·(-1)·2
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*-1}=\frac{12-2\sqrt{38}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*-1}=\frac{12+2\sqrt{38}}{-2}
$