2z^2+1z-2=0

Solution for 2z^2+1z-2=0 equation:

2z^2+1z-2=0

We add all the numbers together, and all the variables

2z^2+z-2=0

a = 2; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·2·(-2)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*2}=\frac{-1-\sqrt{17}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*2}=\frac{-1+\sqrt{17}}{4}
$