1/2z-4=5z-31

Solution for 1/2z-4=5z-31 equation:

1/2z-4=5z-31

We move all terms to the left:

1/2z-4-(5z-31)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We get rid of parentheses

1/2z-5z+31-4=0

We multiply all the terms by the denominator


-5z*2z+31*2z-4*2z+1=0

Wy multiply elements

-10z^2+62z-8z+1=0

We add all the numbers together, and all the variables

-10z^2+54z+1=0

a = -10; b = 54; c = +1;
Δ = b2-4ac
Δ = 542-4·(-10)·1
Δ = 2956
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:

$\sqrt{\Delta}=\sqrt{2956}=\sqrt{4*739}=\sqrt{4}*\sqrt{739}=2\sqrt{739}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-2\sqrt{739}}{2*-10}=\frac{-54-2\sqrt{739}}{-20}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+2\sqrt{739}}{2*-10}=\frac{-54+2\sqrt{739}}{-20}
$