3x^2+6x-797=0

Solution for 3x^2+6x-797=0 equation:

3x^2+6x-797=0

a = 3; b = 6; c = -797;
Δ = b2-4ac
Δ = 62-4·3·(-797)
Δ = 9600
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:

$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-40\sqrt{6}}{2*3}=\frac{-6-40\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+40\sqrt{6}}{2*3}=\frac{-6+40\sqrt{6}}{6}
$