-3x^2-28x+4=0

Solution for -3x^2-28x+4=0 equation:

-3x^2-28x+4=0

a = -3; b = -28; c = +4;
Δ = b2-4ac
Δ = -282-4·(-3)·4
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-8\sqrt{13}}{2*-3}=\frac{28-8\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+8\sqrt{13}}{2*-3}=\frac{28+8\sqrt{13}}{-6}
$