2x^2+64x+1=0

Solution for 2x^2+64x+1=0 equation:

2x^2+64x+1=0

a = 2; b = 64; c = +1;
Δ = b2-4ac
Δ = 642-4·2·1
Δ = 4088
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:

$\sqrt{\Delta}=\sqrt{4088}=\sqrt{4*1022}=\sqrt{4}*\sqrt{1022}=2\sqrt{1022}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-2\sqrt{1022}}{2*2}=\frac{-64-2\sqrt{1022}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+2\sqrt{1022}}{2*2}=\frac{-64+2\sqrt{1022}}{4}
$