2x2+16x=1

Solution for 2x2+16x=1 equation:

2x^2+16x=1

We move all terms to the left:

2x^2+16x-(1)=0

a = 2; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·2·(-1)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{66}}{2*2}=\frac{-16-2\sqrt{66}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{66}}{2*2}=\frac{-16+2\sqrt{66}}{4}
$