-16x^2+98x+3.5=0

Solution for -16x^2+98x+3.5=0 equation:

-16x^2+98x+3.5=0

a = -16; b = 98; c = +3.5;
Δ = b2-4ac
Δ = 982-4·(-16)·3.5
Δ = 9828
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:

$\sqrt{\Delta}=\sqrt{9828}=\sqrt{36*273}=\sqrt{36}*\sqrt{273}=6\sqrt{273}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-6\sqrt{273}}{2*-16}=\frac{-98-6\sqrt{273}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+6\sqrt{273}}{2*-16}=\frac{-98+6\sqrt{273}}{-32}
$