What are the solutions to the equation 8x=4(x^2+2x) ?

The solutions to the equation 8x=4(x^2+2x) are x=0

Find the zeros of 8x=4(x^2+2x)

The zeros of 8x=4(x^2+2x) are x=0

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8 x = 4 (x^{2} + 2 x)

x = 0

Solution steps

8 x = 4 (x^{2} + 2 x)

Move

4 (x^{2} + 2 x)

to the left side

8 x - 4 (x^{2} + 2 x) = 0

Factor

8 x - 4 (x^{2} + 2 x) : - 4 x^{2}

- 4 x^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0

The solution to the quadratic equation is:

x = 0

x^{2} - 2 x + 82

- 3 x^{2} - 5 x = - 1

x, f (x + 1) = 3 x^{2} - x + 2

5 x = - 8 x^{2} + 10

(- (1 - 2) x)^{2} + x^{2} = 1

What are the solutions to the equation 8x=4(x^2+2x) ?
The solutions to the equation 8x=4(x^2+2x) are x=0
Find the zeros of 8x=4(x^2+2x)
The zeros of 8x=4(x^2+2x) are x=0