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

The solutions to the equation -3=x^2+2x-3 are x=0,x=-2

Find the zeros of -3=x^2+2x-3

The zeros of -3=x^2+2x-3 are x=0,x=-2

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

x = 0, x = - 2

Solution steps

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

Switch sides

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

Move

3

to the left side

x^{2} + 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

x = \frac{- 2 + 2}{2 \cdot 1} : 0

x = \frac{- 2 - 2}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

\frac{x}{6} = \frac{1}{2}

\frac{4 x ^{2}}{x ^{2} + 5} < 1

5 x^{2} - 8 x - 4

x (x + 5) = x + 77

1 - \frac{1}{1 - a}

What are the solutions to the equation -3=x^2+2x-3 ?
The solutions to the equation -3=x^2+2x-3 are x=0,x=-2
Find the zeros of -3=x^2+2x-3
The zeros of -3=x^2+2x-3 are x=0,x=-2