What are the solutions to the equation 1=2x^2-x ?

The solutions to the equation 1=2x^2-x are x=1,x=-1/2

Find the zeros of 1=2x^2-x

The zeros of 1=2x^2-x are x=1,x=-1/2

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1 = 2 x^{2} - x

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

Solution steps

1 = 2 x^{2} - x

Switch sides

2 x^{2} - x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 1) = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(x^{- 4})^{3}

\frac{3 ^{2}}{- 4}

0 = 48 t - 16 t^{2}

30 x^{2} - 28 x + 6 = 0

\frac{4 a}{3 b c} - \frac{15 b}{5 a c}

What are the solutions to the equation 1=2x^2-x ?
The solutions to the equation 1=2x^2-x are x=1,x=-1/2
Find the zeros of 1=2x^2-x
The zeros of 1=2x^2-x are x=1,x=-1/2