What are the solutions to the equation -x^2+2x-3=4 ?

The solutions to the equation -x^2+2x-3=4 are x=1-sqrt(6)i,x=1+sqrt(6)i

Find the zeros of -x^2+2x-3=4

The zeros of -x^2+2x-3=4 are x=1-sqrt(6)i,x=1+sqrt(6)i

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- x^{2} + 2 x - 3 = 4

x = 1 - 6 i, x = 1 + 6 i

Solution steps

- x^{2} + 2 x - 3 = 4

Move

4

to the left side

- x^{2} + 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 7 )}{2 ( - 1 )}

Simplify

2^{2} - 4 (- 1) (- 7) : 26 i

x_{1, 2} = \frac{- 2 \pm 2 6 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 6 i}{2 ( - 1 )}, x_{2} = \frac{- 2 - 2 6 i}{2 ( - 1 )}

x = \frac{- 2 + 2 6 i}{2 ( - 1 )} : 1 - 6 i

x = \frac{- 2 - 2 6 i}{2 ( - 1 )} : 1 + 6 i

The solutions to the quadratic equation are:

x = 1 - 6 i, x = 1 + 6 i

(2 x - 1) (3 x - 2) = 3

x, x^{2} + y^{2} + z^{2} + 9 x - 2 y + 10 z + 19 = 0

\frac{n ( n - 3 )}{2} = 4 n

3 x^{2} + 8 x - 14 = 0

\frac{n ( n - 3 )}{2} = 3 n

What are the solutions to the equation -x^2+2x-3=4 ?
The solutions to the equation -x^2+2x-3=4 are x=1-sqrt(6)i,x=1+sqrt(6)i
Find the zeros of -x^2+2x-3=4
The zeros of -x^2+2x-3=4 are x=1-sqrt(6)i,x=1+sqrt(6)i