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

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

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

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

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

x = 2, x = - 1

Solution steps

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

Expand

4 (x + 2) : 4 x + 8

4 x^{2} = 4 x + 8

Move

8

to the left side

4 x^{2} - 8 = 4 x

Move

4 x

to the left side

4 x^{2} - 8 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 ( - 8 )}{2 \cdot 4}

(- 4)^{2} - 4 \cdot 4 (- 8) = 12

x_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 4}

x = \frac{- ( - 4 ) + 12}{2 \cdot 4} : 2

x = \frac{- ( - 4 ) - 12}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

x, x^{2} - x - 12 = 0

z, x^{2} + y^{2} + z^{2} = 1

2 (1 - y^{2}) + y + 1 = 0

(5 x - 1)^{2} - 1 = 15

3 x^{2} - 48 = - 6 x

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