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

The solutions to the equation 4x=x^2+4 are x=2

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4x=x^2+4

4 x = x^{2} + 4

x = 2

Solution steps

4 x = x^{2} + 4

Switch sides

x^{2} + 4 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 4 = 0

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

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

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

x = 2

The solution to the quadratic equation is:

x = 2

x^{2} - 3 x + 14 = 0

- 5 x + 17 = - 5 x^{2}

- 14 = r^{2} - 23

15 x^{2} - 11 x + 2 = 0

5 x^{2} + 2 x + 4 = 0

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