(3/4)z+(1/4)z-6=-5

Solution for (3/4)z+(1/4)z-6=-5 equation:

(3/4)z+(1/4)z-6=-5

We move all terms to the left:

(3/4)z+(1/4)z-6-(-5)=0


Domain of the equation: 4)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+3/4)z+(+1/4)z-6-(-5)=0

We add all the numbers together, and all the variables

(+3/4)z+(+1/4)z-1=0

We multiply parentheses

3z^2+z^2-1=0

We add all the numbers together, and all the variables

4z^2-1=0

a = 4; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·4·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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