What are the solutions to the equation 4(x-1)^2=6x+2 ?

The solutions to the equation 4(x-1)^2=6x+2 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

Find the zeros of 4(x-1)^2=6x+2

The zeros of 4(x-1)^2=6x+2 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

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4 (x - 1)^{2} = 6 x + 2

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

Decimal Notation

x = 3.35078 \dots, x = 0.14921 \dots

Solution steps

4 (x - 1)^{2} = 6 x + 2

Expand

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

4 x^{2} - 8 x + 4 = 6 x + 2

Move

2

to the left side

4 x^{2} - 8 x + 2 = 6 x

Move

6 x

to the left side

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

Solve with the quadratic formula

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

(- 14)^{2} - 4 \cdot 4 \cdot 2 = 241

x_{1, 2} = \frac{- ( - 14 ) \pm 2 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 41}{2 \cdot 4}, x_{2} = \frac{- ( - 14 ) - 2 41}{2 \cdot 4}

x = \frac{- ( - 14 ) + 2 41}{2 \cdot 4} : \frac{7 + 41}{4}

x = \frac{- ( - 14 ) - 2 41}{2 \cdot 4} : \frac{7 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

6 x^{2} + x - 70

\frac{x ^{m}}{x ^{n}}

- 7 x + 6 \leq - 6 (x - 1)

3 x + 7 > 4 x + 2

22 q + 73 > 52 q + 63

What are the solutions to the equation 4(x-1)^2=6x+2 ?
The solutions to the equation 4(x-1)^2=6x+2 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4
Find the zeros of 4(x-1)^2=6x+2
The zeros of 4(x-1)^2=6x+2 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4