5z^2-4z-9=0

Solution for 5z^2-4z-9=0 equation:

5z^2-4z-9=0

a = 5; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·5·(-9)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*5}=\frac{-10}{10}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*5}=\frac{18}{10}
=1+4/5
$