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

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

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

We move all terms to the left:

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

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

We multiply parentheses

5z^2+3z^2-1=0

We add all the numbers together, and all the variables

8z^2-1=0

a = 8; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·8·(-1)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*8}=\frac{0-4\sqrt{2}}{16}
=-\frac{4\sqrt{2}}{16}
=-\frac{\sqrt{2}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*8}=\frac{0+4\sqrt{2}}{16}
=\frac{4\sqrt{2}}{16}
=\frac{\sqrt{2}}{4}
$