What are the solutions to the equation -x^2+4x=5 ?

The solutions to the equation -x^2+4x=5 are x=2-i,x=2+i

Find the zeros of -x^2+4x=5

The zeros of -x^2+4x=5 are x=2-i,x=2+i

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- x^{2} + 4 x = 5

x = 2 - i, x = 2 + i

Solution steps

- x^{2} + 4 x = 5

Move

5

to the left side

- x^{2} + 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

Simplify

4^{2} - 4 (- 1) (- 5) : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 - i, x = 2 + i

- 1 + 8 x \leq - 9

2 x^{2} = 17 x - 41

(x^{2})^{0}

57 + 27

15.3 + 1 h = 1.3 - 1 h

What are the solutions to the equation -x^2+4x=5 ?
The solutions to the equation -x^2+4x=5 are x=2-i,x=2+i
Find the zeros of -x^2+4x=5
The zeros of -x^2+4x=5 are x=2-i,x=2+i