x^2+5x-19x=4

Solution for x^2+5x-19x=4 equation:

x^2+5x-19x=4

We move all terms to the left:

x^2+5x-19x-(4)=0

We add all the numbers together, and all the variables

x^2-14x-4=0

a = 1; b = -14; c = -4;
Δ = b2-4ac
Δ = -142-4·1·(-4)
Δ = 212
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:

$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{53}}{2*1}=\frac{14-2\sqrt{53}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{53}}{2*1}=\frac{14+2\sqrt{53}}{2}
$