3x^2+17x-6=0.

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

3x^2+17x-6=0.

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

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