What are the solutions to the equation 1x^2-2x-3=0 ?

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

Find the zeros of 1x^2-2x-3=0

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

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

x = 3, x = - 1

Solution steps

1 \cdot x^{2} - 2 x - 3 = 0

Refine

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

\frac{x ( x + 1 )}{2} = 91

x^{2} - x - 3

2 x^{2} - 5 x - 3

u^{2} + 16 u

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

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