19x^2+49x-49=0

Solution for 19x^2+49x-49=0 equation:

19x^2+49x-49=0

a = 19; b = 49; c = -49;
Δ = b2-4ac
Δ = 492-4·19·(-49)
Δ = 6125
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:

$\sqrt{\Delta}=\sqrt{6125}=\sqrt{1225*5}=\sqrt{1225}*\sqrt{5}=35\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-35\sqrt{5}}{2*19}=\frac{-49-35\sqrt{5}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+35\sqrt{5}}{2*19}=\frac{-49+35\sqrt{5}}{38}
$