5z-(2z-9)=1/2z+7

Solution for 5z-(2z-9)=1/2z+7 equation:

5z-(2z-9)=1/2z+7

We move all terms to the left:

5z-(2z-9)-(1/2z+7)=0


Domain of the equation: 2z+7)!=0

z∈R


We get rid of parentheses

5z-2z-1/2z+9-7=0

We multiply all the terms by the denominator


5z*2z-2z*2z+9*2z-7*2z-1=0

Wy multiply elements

10z^2-4z^2+18z-14z-1=0

We add all the numbers together, and all the variables

6z^2+4z-1=0

a = 6; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·6·(-1)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*6}=\frac{-4-2\sqrt{10}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*6}=\frac{-4+2\sqrt{10}}{12}
$