-16x^2+64x=2

Solution for -16x^2+64x=2 equation:

-16x^2+64x=2

We move all terms to the left:

-16x^2+64x-(2)=0

a = -16; b = 64; c = -2;
Δ = b2-4ac
Δ = 642-4·(-16)·(-2)
Δ = 3968
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:

$\sqrt{\Delta}=\sqrt{3968}=\sqrt{64*62}=\sqrt{64}*\sqrt{62}=8\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{62}}{2*-16}=\frac{-64-8\sqrt{62}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{62}}{2*-16}=\frac{-64+8\sqrt{62}}{-32}
$