7/2z+5=9/4z+6

Solution for 7/2z+5=9/4z+6 equation:

7/2z+5=9/4z+6

We move all terms to the left:

7/2z+5-(9/4z+6)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: 4z+6)!=0

z∈R


We get rid of parentheses

7/2z-9/4z-6+5=0

We calculate fractions


28z/8z^2+(-18z)/8z^2-6+5=0

We add all the numbers together, and all the variables

28z/8z^2+(-18z)/8z^2-1=0

We multiply all the terms by the denominator


28z+(-18z)-1*8z^2=0

Wy multiply elements

-8z^2+28z+(-18z)=0

We get rid of parentheses

-8z^2+28z-18z=0

We add all the numbers together, and all the variables

-8z^2+10z=0

a = -8; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-8)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-8}=\frac{-20}{-16}
=1+1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-8}=\frac{0}{-16}
=0
$