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

The solutions to the equation -x(-6(x)+1)=1 are x= 1/2 ,x=-1/3

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

The zeros of -x(-6(x)+1)=1 are x= 1/2 ,x=-1/3

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- x (- 6 (x) + 1) = 1

x = \frac{1}{2}, x = - \frac{1}{3}

Decimal Notation

x = 0.5, x = - 0.33333 \dots

Solution steps

- x (- 6 (x) + 1) = 1

Expand

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

6 x^{2} - x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 6 (- 1) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 6}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 6}

x = \frac{- ( - 1 ) + 5}{2 \cdot 6} : \frac{1}{2}

x = \frac{- ( - 1 ) - 5}{2 \cdot 6} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{1}{3}

2 a^{2} + 4 a - 6 = - 3 a

y, 2 x^{2} - x y + y^{2} = 16

y, x^{2} = \frac{1}{y} - 1

x^{2} + \frac{2}{3} x - \frac{1}{9} = 0

4 m^{2} - 15 m + 5 = - 3 m

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