2x^2+4x=16x^2

Solution for 2x^2+4x=16x^2 equation:

2x^2+4x=16x^2

We move all terms to the left:

2x^2+4x-(16x^2)=0

determiningTheFunctionDomain
2x^2-16x^2+4x=0

We add all the numbers together, and all the variables

-14x^2+4x=0

a = -14; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-14)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-14}=\frac{-8}{-28}
=2/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-14}=\frac{0}{-28}
=0
$