What are the solutions to the equation (x^2-x)/(-1)=-3 ?

The solutions to the equation (x^2-x)/(-1)=-3 are x=-(-1+sqrt(13))/2 ,x=(1+sqrt(13))/2

Find the zeros of (x^2-x)/(-1)=-3

The zeros of (x^2-x)/(-1)=-3 are x=-(-1+sqrt(13))/2 ,x=(1+sqrt(13))/2

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

x = - \frac{- 1 + 13}{2}, x = \frac{1 + 13}{2}

Decimal Notation

x = - 1.30277 \dots, x = 2.30277 \dots

Solution steps

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

Expand

\frac{x ^{2} - x}{- 1} : - x^{2} + x

- x^{2} + x = - 3

Move

3

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 (- 1) \cdot 3 = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 ( - 1 )}, x_{2} = \frac{- 1 - 13}{2 ( - 1 )}

x = \frac{- 1 + 13}{2 ( - 1 )} : - \frac{- 1 + 13}{2}

x = \frac{- 1 - 13}{2 ( - 1 )} : \frac{1 + 13}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 13}{2}, x = \frac{1 + 13}{2}

6 x^{2} + x - 35 = 0

4 b^{2} - 4 b + 7 = 66

r, 0.0119 = \frac{1}{2} (326) r^{2}

6.28 (\frac{4 - x}{2})^{2} = 20.12

(x - 4) (x - 10) = 0

What are the solutions to the equation (x^2-x)/(-1)=-3 ?
The solutions to the equation (x^2-x)/(-1)=-3 are x=-(-1+sqrt(13))/2 ,x=(1+sqrt(13))/2
Find the zeros of (x^2-x)/(-1)=-3
The zeros of (x^2-x)/(-1)=-3 are x=-(-1+sqrt(13))/2 ,x=(1+sqrt(13))/2