What are the solutions to the equation (4x-2)^2-4(4x-2)+3=0 ?

The solutions to the equation (4x-2)^2-4(4x-2)+3=0 are x= 5/4 ,x= 3/4

Find the zeros of (4x-2)^2-4(4x-2)+3=0

The zeros of (4x-2)^2-4(4x-2)+3=0 are x= 5/4 ,x= 3/4

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(4 x - 2)^{2} - 4 (4 x - 2) + 3 = 0

x = \frac{5}{4}, x = \frac{3}{4}

Decimal

x = 1.25, x = 0.75

Solution steps

(4 x - 2)^{2} - 4 (4 x - 2) + 3 = 0

Expand

(4 x - 2)^{2} - 4 (4 x - 2) + 3 : 16 x^{2} - 32 x + 15

16 x^{2} - 32 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 16 \cdot 15}{2 \cdot 16}

(- 32)^{2} - 4 \cdot 16 \cdot 15 = 8

x_{1, 2} = \frac{- ( - 32 ) \pm 8}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 8}{2 \cdot 16}, x_{2} = \frac{- ( - 32 ) - 8}{2 \cdot 16}

x = \frac{- ( - 32 ) + 8}{2 \cdot 16} : \frac{5}{4}

x = \frac{- ( - 32 ) - 8}{2 \cdot 16} : \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4}, x = \frac{3}{4}

- 1 - (- 5)

\frac{6}{6 + 1} - 1

x^{2} + 3 x - 4 \leq 0

3^{5} < 4^{6}

6 x + 11 + 11 x + 16 = 180

What are the solutions to the equation (4x-2)^2-4(4x-2)+3=0 ?
The solutions to the equation (4x-2)^2-4(4x-2)+3=0 are x= 5/4 ,x= 3/4
Find the zeros of (4x-2)^2-4(4x-2)+3=0
The zeros of (4x-2)^2-4(4x-2)+3=0 are x= 5/4 ,x= 3/4