-16x^2+160x+6=0

Solution for -16x^2+160x+6=0 equation:

-16x^2+160x+6=0

a = -16; b = 160; c = +6;
Δ = b2-4ac
Δ = 1602-4·(-16)·6
Δ = 25984
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:

$\sqrt{\Delta}=\sqrt{25984}=\sqrt{64*406}=\sqrt{64}*\sqrt{406}=8\sqrt{406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{406}}{2*-16}=\frac{-160-8\sqrt{406}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{406}}{2*-16}=\frac{-160+8\sqrt{406}}{-32}
$