What are the solutions to the equation 2(4x+1)(x-1)+3=0 ?

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

Find the zeros of 2(4x+1)(x-1)+3=0

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

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2 (4 x + 1) (x - 1) + 3 = 0

x = \frac{1}{2}, x = \frac{1}{4}

Decimal Notation

x = 0.5, x = 0.25

Solution steps

2 (4 x + 1) (x - 1) + 3 = 0

Expand

2 (4 x + 1) (x - 1) + 3 : 8 x^{2} - 6 x + 1

8 x^{2} - 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

(- 6)^{2} - 4 \cdot 8 \cdot 1 = 2

x_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 8}, x_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 8}

x = \frac{- ( - 6 ) + 2}{2 \cdot 8} : \frac{1}{2}

x = \frac{- ( - 6 ) - 2}{2 \cdot 8} : \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = \frac{1}{4}

\frac{x ^{2}}{3} = 12

\frac{x - 5}{4 x - 4} + \frac{2}{x ^{2} - 1} \geq 0

12 n^{2} - 7 n + 1

69 = 1 \cdot 0.037 \cdot (x - 23.45)

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