z^2-3z-7/4=0

Solution for z^2-3z-7/4=0 equation:

z^2-3z-7/4=0

We multiply all the terms by the denominator


z^2*4-3z*4-7=0

Wy multiply elements

4z^2-12z-7=0

a = 4; b = -12; c = -7;
Δ = b2-4ac
Δ = -122-4·4·(-7)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*4}=\frac{-4}{8}
=-1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*4}=\frac{28}{8}
=3+1/2
$