6+7/4z=24+z

Solution for 6+7/4z=24+z equation:

6+7/4z=24+z

We move all terms to the left:

6+7/4z-(24+z)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R


We add all the numbers together, and all the variables

7/4z-(z+24)+6=0

We get rid of parentheses

7/4z-z-24+6=0

We multiply all the terms by the denominator


-z*4z-24*4z+6*4z+7=0

Wy multiply elements

-4z^2-96z+24z+7=0

We add all the numbers together, and all the variables

-4z^2-72z+7=0

a = -4; b = -72; c = +7;
Δ = b2-4ac
Δ = -722-4·(-4)·7
Δ = 5296
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:

$\sqrt{\Delta}=\sqrt{5296}=\sqrt{16*331}=\sqrt{16}*\sqrt{331}=4\sqrt{331}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-4\sqrt{331}}{2*-4}=\frac{72-4\sqrt{331}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+4\sqrt{331}}{2*-4}=\frac{72+4\sqrt{331}}{-8}
$