What are the solutions to the equation x^2-2x=-3/6 ?

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

Find the zeros of x^2-2x=-3/6

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

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x^{2} - 2 x = - \frac{3}{6}

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

Solution steps

x^{2} - 2 x = - \frac{3}{6}

Multiply both sides by

6

6 x^{2} - 12 x = - 3

Move

3

to the left side

6 x^{2} - 12 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

(- 12)^{2} - 4 \cdot 6 \cdot 3 = 62

x_{1, 2} = \frac{- ( - 12 ) \pm 6 2}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6 2}{2 \cdot 6}, x_{2} = \frac{- ( - 12 ) - 6 2}{2 \cdot 6}

x = \frac{- ( - 12 ) + 6 2}{2 \cdot 6} : \frac{2 + 2}{2}

x = \frac{- ( - 12 ) - 6 2}{2 \cdot 6} : \frac{2 - 2}{2}

The solutions to the quadratic equation are:

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

146

x^{4} - 6 x^{2} - 16

8 a (x + y) - (x + y)

7 x^{2} + 6 x + 1 = 0

\frac{y ^{3}}{y ^{2}}

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