3x^2+2x-634=0

Solution for 3x^2+2x-634=0 equation:

3x^2+2x-634=0

a = 3; b = 2; c = -634;
Δ = b2-4ac
Δ = 22-4·3·(-634)
Δ = 7612
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:

$\sqrt{\Delta}=\sqrt{7612}=\sqrt{4*1903}=\sqrt{4}*\sqrt{1903}=2\sqrt{1903}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1903}}{2*3}=\frac{-2-2\sqrt{1903}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1903}}{2*3}=\frac{-2+2\sqrt{1903}}{6}
$