16x^2+96x+5=0

Solution for 16x^2+96x+5=0 equation:

16x^2+96x+5=0

a = 16; b = 96; c = +5;
Δ = b2-4ac
Δ = 962-4·16·5
Δ = 8896
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:

$\sqrt{\Delta}=\sqrt{8896}=\sqrt{64*139}=\sqrt{64}*\sqrt{139}=8\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{139}}{2*16}=\frac{-96-8\sqrt{139}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{139}}{2*16}=\frac{-96+8\sqrt{139}}{32}
$