What are the solutions to the equation 0=x^2+2x-3 ?

The solutions to the equation 0=x^2+2x-3 are x=1,x=-3

Find the zeros of 0=x^2+2x-3

The zeros of 0=x^2+2x-3 are x=1,x=-3

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

x = 1, x = - 3

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- 2 + 4}{2 \cdot 1}, x_{2} = \frac{- 2 - 4}{2 \cdot 1}

x = \frac{- 2 + 4}{2 \cdot 1} : 1

x = \frac{- 2 - 4}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

\frac{2 ( 1 )}{1 + 4}

53

\frac{4 x + 8}{5} < 4

\frac{x ^{7} y ^{- 4}}{x ^{- 2} y ^{- 6}}

12 + 48

What are the solutions to the equation 0=x^2+2x-3 ?
The solutions to the equation 0=x^2+2x-3 are x=1,x=-3
Find the zeros of 0=x^2+2x-3
The zeros of 0=x^2+2x-3 are x=1,x=-3