19x2+3x-9=0

Solution for 19x2+3x-9=0 equation:

19x^2+3x-9=0

a = 19; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·19·(-9)
Δ = 693
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{693}=\sqrt{9*77}=\sqrt{9}*\sqrt{77}=3\sqrt{77}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{77}}{2*19}=\frac{-3-3\sqrt{77}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{77}}{2*19}=\frac{-3+3\sqrt{77}}{38}
$