What are the solutions to the equation -x^2+4x=13 ?

The solutions to the equation -x^2+4x=13 are x=2-3i,x=2+3i

Find the zeros of -x^2+4x=13

The zeros of -x^2+4x=13 are x=2-3i,x=2+3i

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- x^{2} + 4 x = 13

x = 2 - 3 i, x = 2 + 3 i

Solution steps

- x^{2} + 4 x = 13

Move

13

to the left side

- x^{2} + 4 x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) ( - 13 )}{2 ( - 1 )}

Simplify

4^{2} - 4 (- 1) (- 13) : 6 i

x_{1, 2} = \frac{- 4 \pm 6 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 6 i}{2 ( - 1 )}, x_{2} = \frac{- 4 - 6 i}{2 ( - 1 )}

x = \frac{- 4 + 6 i}{2 ( - 1 )} : 2 - 3 i

x = \frac{- 4 - 6 i}{2 ( - 1 )} : 2 + 3 i

The solutions to the quadratic equation are:

x = 2 - 3 i, x = 2 + 3 i

15 = \frac{a}{3} - 2

2 y - 9 = y - 2

3 x^{2} - 5 x - 2 \geq 0

\frac{( 1 - s ^{4} )}{1 + s ^{4}}

(x - 8) (x + 2) = - 24

What are the solutions to the equation -x^2+4x=13 ?
The solutions to the equation -x^2+4x=13 are x=2-3i,x=2+3i
Find the zeros of -x^2+4x=13
The zeros of -x^2+4x=13 are x=2-3i,x=2+3i