What are the solutions to the equation 3(4x-1)^2-7(4x-1)=-4 ?

The solutions to the equation 3(4x-1)^2-7(4x-1)=-4 are x= 7/12 ,x= 1/2

Find the zeros of 3(4x-1)^2-7(4x-1)=-4

The zeros of 3(4x-1)^2-7(4x-1)=-4 are x= 7/12 ,x= 1/2

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3 (4 x - 1)^{2} - 7 (4 x - 1) = - 4

x = \frac{7}{12}, x = \frac{1}{2}

Decimal Notation

x = 0.58333 \dots, x = 0.5

Solution steps

3 (4 x - 1)^{2} - 7 (4 x - 1) = - 4

Expand

3 (4 x - 1)^{2} - 7 (4 x - 1) : 48 x^{2} - 52 x + 10

48 x^{2} - 52 x + 10 = - 4

Move

4

to the left side

48 x^{2} - 52 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 52 ) \pm ( - 52 ) ^{2} - 4 \cdot 48 \cdot 14}{2 \cdot 48}

(- 52)^{2} - 4 \cdot 48 \cdot 14 = 4

x_{1, 2} = \frac{- ( - 52 ) \pm 4}{2 \cdot 48}

Separate the solutions

x_{1} = \frac{- ( - 52 ) + 4}{2 \cdot 48}, x_{2} = \frac{- ( - 52 ) - 4}{2 \cdot 48}

x = \frac{- ( - 52 ) + 4}{2 \cdot 48} : \frac{7}{12}

x = \frac{- ( - 52 ) - 4}{2 \cdot 48} : \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{7}{12}, x = \frac{1}{2}

4 - 2 x < 3 (3 - x)

X^{4} - Y^{4}

3 x^{2} + x - 6 = 0

2 x^{2} - 13 x + 14 = 0

192 x^{2} - 243 y^{3}

What are the solutions to the equation 3(4x-1)^2-7(4x-1)=-4 ?
The solutions to the equation 3(4x-1)^2-7(4x-1)=-4 are x= 7/12 ,x= 1/2
Find the zeros of 3(4x-1)^2-7(4x-1)=-4
The zeros of 3(4x-1)^2-7(4x-1)=-4 are x= 7/12 ,x= 1/2