(9/4)z+(1/2)=2

Solution for (9/4)z+(1/2)=2 equation:

(9/4)z+(1/2)=2

We move all terms to the left:

(9/4)z+(1/2)-(2)=0


Domain of the equation: 4)z!=0

z!=0/1

z!=0

z∈R


determiningTheFunctionDomain
(9/4)z-2+(1/2)=0

We add all the numbers together, and all the variables

(+9/4)z-2+(+1/2)=0

We multiply parentheses

9z^2-2+(+1/2)=0

We get rid of parentheses

9z^2-2+1/2=0

We multiply all the terms by the denominator


9z^2*2+1-2*2=0

We add all the numbers together, and all the variables

9z^2*2-3=0

Wy multiply elements

18z^2-3=0

a = 18; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·18·(-3)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*18}=\frac{0-6\sqrt{6}}{36}
=-\frac{6\sqrt{6}}{36}
=-\frac{\sqrt{6}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*18}=\frac{0+6\sqrt{6}}{36}
=\frac{6\sqrt{6}}{36}
=\frac{\sqrt{6}}{6}
$