-6z^2+17z+3=0

Solution for -6z^2+17z+3=0 equation:

-6z^2+17z+3=0

a = -6; b = 17; c = +3;
Δ = b2-4ac
Δ = 172-4·(-6)·3
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*-6}=\frac{-36}{-12}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*-6}=\frac{2}{-12}
=-1/6
$