/6z+10=4z+19

Solution for /6z+10=4z+19 equation:

/6z+10=4z+19

We move all terms to the left:

/6z+10-(4z+19)=0


Domain of the equation: 6z!=0

z!=0/6

z!=0

z∈R


We get rid of parentheses

/6z-4z-19+10=0

We multiply all the terms by the denominator


-4z*6z-19*6z+10*6z+=0

We add all the numbers together, and all the variables

-4z*6z-19*6z+10*6z=0

Wy multiply elements

-24z^2-114z+60z=0

We add all the numbers together, and all the variables

-24z^2-54z=0

a = -24; b = -54; c = 0;
Δ = b2-4ac
Δ = -542-4·(-24)·0
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-54}{2*-24}=\frac{0}{-48}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+54}{2*-24}=\frac{108}{-48}
=-2+1/4
$