3x^2+14x-196=0

Solution for 3x^2+14x-196=0 equation:

3x^2+14x-196=0

a = 3; b = 14; c = -196;
Δ = b2-4ac
Δ = 142-4·3·(-196)
Δ = 2548
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:

$\sqrt{\Delta}=\sqrt{2548}=\sqrt{196*13}=\sqrt{196}*\sqrt{13}=14\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{13}}{2*3}=\frac{-14-14\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{13}}{2*3}=\frac{-14+14\sqrt{13}}{6}
$