3x^2+2x-325=0

Solution for 3x^2+2x-325=0 equation:

3x^2+2x-325=0

a = 3; b = 2; c = -325;
Δ = b2-4ac
Δ = 22-4·3·(-325)
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8\sqrt{61}}{2*3}=\frac{-2-8\sqrt{61}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8\sqrt{61}}{2*3}=\frac{-2+8\sqrt{61}}{6}
$