2x^2-4x=47

Solution for 2x^2-4x=47 equation:

2x^2-4x=47

We move all terms to the left:

2x^2-4x-(47)=0

a = 2; b = -4; c = -47;
Δ = b2-4ac
Δ = -42-4·2·(-47)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14\sqrt{2}}{2*2}=\frac{4-14\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14\sqrt{2}}{2*2}=\frac{4+14\sqrt{2}}{4}
$