15x^2+19x-56=0

Solution for 15x^2+19x-56=0 equation:

15x^2+19x-56=0

a = 15; b = 19; c = -56;
Δ = b2-4ac
Δ = 192-4·15·(-56)
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-61}{2*15}=\frac{-80}{30}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+61}{2*15}=\frac{42}{30}
=1+2/5
$