What are the solutions to the equation 6x^2-8x+1=x+4 ?

The solutions to the equation 6x^2-8x+1=x+4 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4

Find the zeros of 6x^2-8x+1=x+4

The zeros of 6x^2-8x+1=x+4 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4

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6 x^{2} - 8 x + 1 = x + 4

x = \frac{3 + 17}{4}, x = \frac{3 - 17}{4}

Decimal Notation

x = 1.78077 \dots, x = - 0.28077 \dots

Solution steps

6 x^{2} - 8 x + 1 = x + 4

Move

4

to the left side

6 x^{2} - 8 x - 3 = x

Move

x

to the left side

6 x^{2} - 9 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 9)^{2} - 4 \cdot 6 (- 3) = 317

x_{1, 2} = \frac{- ( - 9 ) \pm 3 17}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3 17}{2 \cdot 6}, x_{2} = \frac{- ( - 9 ) - 3 17}{2 \cdot 6}

x = \frac{- ( - 9 ) + 3 17}{2 \cdot 6} : \frac{3 + 17}{4}

x = \frac{- ( - 9 ) - 3 17}{2 \cdot 6} : \frac{3 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 17}{4}, x = \frac{3 - 17}{4}

(4 x - 2) (- 2 x + 5) = 0

- 12 t^{2} + 122 t - 109 = 0

5 c^{2} + 12 c + 4 = 0

2 (x - 5)^{2} + 2 = 20

8 x^{2} = 72 x

What are the solutions to the equation 6x^2-8x+1=x+4 ?
The solutions to the equation 6x^2-8x+1=x+4 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4
Find the zeros of 6x^2-8x+1=x+4
The zeros of 6x^2-8x+1=x+4 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4