16x^2-96x+139=0

Solution for 16x^2-96x+139=0 equation:

16x^2-96x+139=0

a = 16; b = -96; c = +139;
Δ = b2-4ac
Δ = -962-4·16·139
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{5}}{2*16}=\frac{96-8\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{5}}{2*16}=\frac{96+8\sqrt{5}}{32}
$