16x^2-112x+98=0

Solution for 16x^2-112x+98=0 equation:

16x^2-112x+98=0

a = 16; b = -112; c = +98;
Δ = b2-4ac
Δ = -1122-4·16·98
Δ = 6272
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:

$\sqrt{\Delta}=\sqrt{6272}=\sqrt{3136*2}=\sqrt{3136}*\sqrt{2}=56\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-56\sqrt{2}}{2*16}=\frac{112-56\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+56\sqrt{2}}{2*16}=\frac{112+56\sqrt{2}}{32}
$