6x2+19x+-3=0

Solution for 6x2+19x+-3=0 equation:

6x^2+19x+-3=0

We add all the numbers together, and all the variables

6x^2+19x=0

a = 6; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·6·0
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{-(19)-19}{2*6}=\frac{-38}{12}
=-3+1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*6}=\frac{0}{12}
=0
$