3x^2+4x-2308=0

Solution for 3x^2+4x-2308=0 equation:

3x^2+4x-2308=0

a = 3; b = 4; c = -2308;
Δ = b2-4ac
Δ = 42-4·3·(-2308)
Δ = 27712
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:

$\sqrt{\Delta}=\sqrt{27712}=\sqrt{64*433}=\sqrt{64}*\sqrt{433}=8\sqrt{433}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{433}}{2*3}=\frac{-4-8\sqrt{433}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{433}}{2*3}=\frac{-4+8\sqrt{433}}{6}
$