-3x^2+12x+429=0

Solution for -3x^2+12x+429=0 equation:

-3x^2+12x+429=0

a = -3; b = 12; c = +429;
Δ = b2-4ac
Δ = 122-4·(-3)·429
Δ = 5292
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:

$\sqrt{\Delta}=\sqrt{5292}=\sqrt{1764*3}=\sqrt{1764}*\sqrt{3}=42\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-42\sqrt{3}}{2*-3}=\frac{-12-42\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+42\sqrt{3}}{2*-3}=\frac{-12+42\sqrt{3}}{-6}
$