-16x^2+224x+168=0

Solution for -16x^2+224x+168=0 equation:

-16x^2+224x+168=0

a = -16; b = 224; c = +168;
Δ = b2-4ac
Δ = 2242-4·(-16)·168
Δ = 60928
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:

$\sqrt{\Delta}=\sqrt{60928}=\sqrt{256*238}=\sqrt{256}*\sqrt{238}=16\sqrt{238}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(224)-16\sqrt{238}}{2*-16}=\frac{-224-16\sqrt{238}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(224)+16\sqrt{238}}{2*-16}=\frac{-224+16\sqrt{238}}{-32}
$