What are the solutions to the equation x^2+4x=1(16+4x) ?

The solutions to the equation x^2+4x=1(16+4x) are x=4,x=-4

Find the zeros of x^2+4x=1(16+4x)

The zeros of x^2+4x=1(16+4x) are x=4,x=-4

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x^{2} + 4 x = 1 (16 + 4 x)

x = 4, x = - 4

Solution steps

x^{2} + 4 x = 1 \cdot (16 + 4 x)

Expand

1 \cdot (16 + 4 x) : 16 + 4 x

x^{2} + 4 x = 16 + 4 x

Move

4 x

to the left side

x^{2} = 16

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 16, x = - 16

16 = 4

- 16 = - 4

x = 4, x = - 4

3 x^{2} + 6 x + 4 x + 8

2125

\frac{9 i}{- 4 + 2 i}

4 y^{2} = - 12 y

\frac{2 ( \frac{8}{15} )}{1 - ( \frac{8}{15} ) ^{2}}

What are the solutions to the equation x^2+4x=1(16+4x) ?
The solutions to the equation x^2+4x=1(16+4x) are x=4,x=-4
Find the zeros of x^2+4x=1(16+4x)
The zeros of x^2+4x=1(16+4x) are x=4,x=-4