16^3-28x^2+12x=0

Solution for 16^3-28x^2+12x=0 equation:

16^3-28x^2+12x=0

We add all the numbers together, and all the variables

-28x^2+12x+4096=0

a = -28; b = 12; c = +4096;
Δ = b2-4ac
Δ = 122-4·(-28)·4096
Δ = 458896
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:

$\sqrt{\Delta}=\sqrt{458896}=\sqrt{16*28681}=\sqrt{16}*\sqrt{28681}=4\sqrt{28681}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{28681}}{2*-28}=\frac{-12-4\sqrt{28681}}{-56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{28681}}{2*-28}=\frac{-12+4\sqrt{28681}}{-56}
$