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

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

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

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

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

x = 3, x = - 1

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} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 1

4 x^{2} - 2 x - 12 = 0

- 15 \cdot - 5

(\frac{3 t ^{6} u ^{2} v ^{5}}{9 t u v ^{21}})^{0}

4^{- x}

4 x^{2} + 25 = 0

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