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

Solution for -2z^2-5z+3=0 equation:

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

a = -2; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·(-2)·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*-2}=\frac{-2}{-4}
=1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*-2}=\frac{12}{-4}
=-3
$