5-3/4z=5/6z+10

Solution for 5-3/4z=5/6z+10 equation:

5-3/4z=5/6z+10

We move all terms to the left:

5-3/4z-(5/6z+10)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 6z+10)!=0

z∈R


We get rid of parentheses

-3/4z-5/6z-10+5=0

We calculate fractions


(-18z)/24z^2+(-20z)/24z^2-10+5=0

We add all the numbers together, and all the variables

(-18z)/24z^2+(-20z)/24z^2-5=0

We multiply all the terms by the denominator


(-18z)+(-20z)-5*24z^2=0

Wy multiply elements

-120z^2+(-18z)+(-20z)=0

We get rid of parentheses

-120z^2-18z-20z=0

We add all the numbers together, and all the variables

-120z^2-38z=0

a = -120; b = -38; c = 0;
Δ = b2-4ac
Δ = -382-4·(-120)·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-38}{2*-120}=\frac{0}{-240}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+38}{2*-120}=\frac{76}{-240}
=-19/60
$