z^2=z+3/4

Solution for z^2=z+3/4 equation:

z^2=z+3/4

We move all terms to the left:

z^2-(z+3/4)=0

We add all the numbers together, and all the variables

z^2-(+z+3/4)=0

We get rid of parentheses

z^2-z-3/4=0

We multiply all the terms by the denominator


z^2*4-z*4-3=0

Wy multiply elements

4z^2-4z-3=0

a = 4; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·4·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*4}=\frac{-4}{8}
=-1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*4}=\frac{12}{8}
=1+1/2
$