What are the solutions to the equation 3x^2-6x+1=-4x+6 ?

The solutions to the equation 3x^2-6x+1=-4x+6 are x= 5/3 ,x=-1

Find the zeros of 3x^2-6x+1=-4x+6

The zeros of 3x^2-6x+1=-4x+6 are x= 5/3 ,x=-1

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

x = \frac{5}{3}, x = - 1

Decimal Notation

x = 1.66666 \dots, x = - 1

Solution steps

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

Move

6

to the left side

3 x^{2} - 6 x - 5 = - 4 x

Move

4 x

to the left side

3 x^{2} - 2 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 5) = 8

x_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 3}

x = \frac{- ( - 2 ) + 8}{2 \cdot 3} : \frac{5}{3}

x = \frac{- ( - 2 ) - 8}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = \frac{5}{3}, x = - 1

x^{2} - 16 < 0

\frac{3}{5} + \frac{1}{7} x - \frac{1}{3} = \frac{6}{7} x

7 x - 3 (5 x - 4) = 20

y^{3} - 3 y^{2} + 4 y - 12

- 5 + 7

What are the solutions to the equation 3x^2-6x+1=-4x+6 ?
The solutions to the equation 3x^2-6x+1=-4x+6 are x= 5/3 ,x=-1
Find the zeros of 3x^2-6x+1=-4x+6
The zeros of 3x^2-6x+1=-4x+6 are x= 5/3 ,x=-1