19x2+10x-4=0

Solution for 19x2+10x-4=0 equation:

19x^2+10x-4=0

a = 19; b = 10; c = -4;
Δ = b2-4ac
Δ = 102-4·19·(-4)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{101}}{2*19}=\frac{-10-2\sqrt{101}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{101}}{2*19}=\frac{-10+2\sqrt{101}}{38}
$