3x^2+28x-233=0

Solution for 3x^2+28x-233=0 equation:

3x^2+28x-233=0

a = 3; b = 28; c = -233;
Δ = b2-4ac
Δ = 282-4·3·(-233)
Δ = 3580
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:

$\sqrt{\Delta}=\sqrt{3580}=\sqrt{4*895}=\sqrt{4}*\sqrt{895}=2\sqrt{895}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{895}}{2*3}=\frac{-28-2\sqrt{895}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{895}}{2*3}=\frac{-28+2\sqrt{895}}{6}
$