-16x^2+62x+2=0

Solution for -16x^2+62x+2=0 equation:

-16x^2+62x+2=0

a = -16; b = 62; c = +2;
Δ = b2-4ac
Δ = 622-4·(-16)·2
Δ = 3972
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:

$\sqrt{\Delta}=\sqrt{3972}=\sqrt{4*993}=\sqrt{4}*\sqrt{993}=2\sqrt{993}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-2\sqrt{993}}{2*-16}=\frac{-62-2\sqrt{993}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+2\sqrt{993}}{2*-16}=\frac{-62+2\sqrt{993}}{-32}
$