47=2x2+4x

Solution for 47=2x2+4x equation:

47=2x^2+4x

We move all terms to the left:

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

We get rid of parentheses

-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}
$