15x^2+-19x+6=0

Solution for 15x^2+-19x+6=0 equation:

15x^2+-19x+6=0

We add all the numbers together, and all the variables

15x^2-19x=0

a = 15; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·15·0
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*15}=\frac{0}{30}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*15}=\frac{38}{30}
=1+4/15
$