What are the solutions to the equation (4x-1)^2-4(4x-1)+4=0 ?

The solutions to the equation (4x-1)^2-4(4x-1)+4=0 are x= 3/4

Find the zeros of (4x-1)^2-4(4x-1)+4=0

The zeros of (4x-1)^2-4(4x-1)+4=0 are x= 3/4

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(4 x - 1)^{2} - 4 (4 x - 1) + 4 = 0

x = \frac{3}{4}

Decimal Notation

x = 0.75

Solution steps

(4 x - 1)^{2} - 4 (4 x - 1) + 4 = 0

Expand

(4 x - 1)^{2} - 4 (4 x - 1) + 4 : 16 x^{2} - 24 x + 9

16 x^{2} - 24 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 16 \cdot 9}{2 \cdot 16}

(- 24)^{2} - 4 \cdot 16 \cdot 9 = 0

x_{1, 2} = \frac{- ( - 24 ) \pm 0}{2 \cdot 16}

x = \frac{- ( - 24 )}{2 \cdot 16}

\frac{- ( - 24 )}{2 \cdot 16} = \frac{3}{4}

x = \frac{3}{4}

The solution to the quadratic equation is:

x = \frac{3}{4}

2 x^{2} = 10 x + 5

(x^{2} + x - 6) = 0

3 x^{2} - x + 12 = 0

x^{2} - 10 x = 4

n^{2} - 7 n = - 10

What are the solutions to the equation (4x-1)^2-4(4x-1)+4=0 ?
The solutions to the equation (4x-1)^2-4(4x-1)+4=0 are x= 3/4
Find the zeros of (4x-1)^2-4(4x-1)+4=0
The zeros of (4x-1)^2-4(4x-1)+4=0 are x= 3/4