80x-16x^2+3.5=0

Solution for 80x-16x^2+3.5=0 equation:

80x-16x^2+3.5=0

a = -16; b = 80; c = +3.5;
Δ = b2-4ac
Δ = 802-4·(-16)·3.5
Δ = 6624
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:

$\sqrt{\Delta}=\sqrt{6624}=\sqrt{144*46}=\sqrt{144}*\sqrt{46}=12\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-12\sqrt{46}}{2*-16}=\frac{-80-12\sqrt{46}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+12\sqrt{46}}{2*-16}=\frac{-80+12\sqrt{46}}{-32}
$