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

The solutions to the equation -2=-x+x^2-4 are x=2,x=-1

Find the zeros of -2=-x+x^2-4

The zeros of -2=-x+x^2-4 are x=2,x=-1

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- 2 = - x + x^{2} - 4

x = 2, x = - 1

Solution steps

- 2 = - x + x^{2} - 4

Switch sides

- x + x^{2} - 4 = - 2

Move

2

to the left side

x^{2} - x - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = - 1

x^{2} - 6 x + 8 > 0

\frac{3}{5} x \leq 4

x^{2} - x - 1 < 0

e^{4 t}

3 x^{2} + 25 x + 42 = 0

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