1/2z-4=3z-19

Solution for 1/2z-4=3z-19 equation:

1/2z-4=3z-19

We move all terms to the left:

1/2z-4-(3z-19)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We get rid of parentheses

1/2z-3z+19-4=0

We multiply all the terms by the denominator


-3z*2z+19*2z-4*2z+1=0

Wy multiply elements

-6z^2+38z-8z+1=0

We add all the numbers together, and all the variables

-6z^2+30z+1=0

a = -6; b = 30; c = +1;
Δ = b2-4ac
Δ = 302-4·(-6)·1
Δ = 924
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:

$\sqrt{\Delta}=\sqrt{924}=\sqrt{4*231}=\sqrt{4}*\sqrt{231}=2\sqrt{231}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{231}}{2*-6}=\frac{-30-2\sqrt{231}}{-12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{231}}{2*-6}=\frac{-30+2\sqrt{231}}{-12}
$