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

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

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

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

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2 x - x^{2} = - 3

x = - 1, x = 3

Solution steps

2 x - x^{2} = - 3

Move

3

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

- 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

y^{2} - 81

6 t - 6 t^{2} = 0

x + 3 \geq 4

\frac{7}{3} + 5

(x - 5)^{2} = 0

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