What are the solutions to the equation 4x^2-16x+16=0 ?

The solutions to the equation 4x^2-16x+16=0 are x=2

Find the zeros of 4x^2-16x+16=0

The zeros of 4x^2-16x+16=0 are x=2

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4 x^{2} - 16 x + 16 = 0

x = 2

Solution steps

4 x^{2} - 16 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 16 = 0

x_{1, 2} = \frac{- ( - 16 ) \pm 0}{2 \cdot 4}

x = \frac{- ( - 16 )}{2 \cdot 4}

\frac{- ( - 16 )}{2 \cdot 4} = 2

x = 2

The solution to the quadratic equation is:

x = 2

\frac{7}{10} x - \frac{1}{5} = 4

\frac{40 a}{40 a ^{2}}

(\frac{8 b ^{4}}{4 b ^{2}})^{3}

(3 x + 3) (x + 2) = 1

25 b^{4} - 36 b^{2}

What are the solutions to the equation 4x^2-16x+16=0 ?
The solutions to the equation 4x^2-16x+16=0 are x=2
Find the zeros of 4x^2-16x+16=0
The zeros of 4x^2-16x+16=0 are x=2