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

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

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

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

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

x = - 1, x = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = 3

x^{2} + 4 x + 5 < 0

(3 x y^{3})^{2} (x y)^{6}

2 x^{2} - 5 x - 10 = 0

2 x^{2} + 5 x < - x^{2} + 1

(x + 2) = 0

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