4/2z+6=10/7z-24

Solution for 4/2z+6=10/7z-24 equation:

4/2z+6=10/7z-24

We move all terms to the left:

4/2z+6-(10/7z-24)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: 7z-24)!=0

z∈R


We get rid of parentheses

4/2z-10/7z+24+6=0

We calculate fractions


28z/14z^2+(-20z)/14z^2+24+6=0

We add all the numbers together, and all the variables

28z/14z^2+(-20z)/14z^2+30=0

We multiply all the terms by the denominator


28z+(-20z)+30*14z^2=0

Wy multiply elements

420z^2+28z+(-20z)=0

We get rid of parentheses

420z^2+28z-20z=0

We add all the numbers together, and all the variables

420z^2+8z=0

a = 420; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·420·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*420}=\frac{-16}{840}
=-2/105
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*420}=\frac{0}{840}
=0
$