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

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

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-4=x^2-4x

- 4 = x^{2} - 4 x

x = 2

Solution steps

- 4 = x^{2} - 4 x

Switch sides

x^{2} - 4 x = - 4

Move

4

to the left side

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

6 x^{2} + 8 x

x^{2} + x - \frac{1}{4} = 0

r^{2} + r + 6 = 0

y, (x - 3)^{2} + (y + 2)^{2} = 4

(2 s - 1)^{2} = 225

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