16x+2-2x^2=0

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

16x+2-2x^2=0

a = -2; b = 16; c = +2;
Δ = b2-4ac
Δ = 162-4·(-2)·2
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{17}}{2*-2}=\frac{-16-4\sqrt{17}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{17}}{2*-2}=\frac{-16+4\sqrt{17}}{-4}
$