4x-3x^2+124=0

Solution for 4x-3x^2+124=0 equation:

4x-3x^2+124=0

a = -3; b = 4; c = +124;
Δ = b2-4ac
Δ = 42-4·(-3)·124
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{94}}{2*-3}=\frac{-4-4\sqrt{94}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{94}}{2*-3}=\frac{-4+4\sqrt{94}}{-6}
$