What are the solutions to the equation -x^2+9x=4x+36 ?

The solutions to the equation -x^2+9x=4x+36 are x= 5/2-i(sqrt(119))/2 ,x= 5/2+i(sqrt(119))/2

Find the zeros of -x^2+9x=4x+36

The zeros of -x^2+9x=4x+36 are x= 5/2-i(sqrt(119))/2 ,x= 5/2+i(sqrt(119))/2

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- x^{2} + 9 x = 4 x + 36

x = \frac{5}{2} - i \frac{119}{2}, x = \frac{5}{2} + i \frac{119}{2}

Solution steps

- x^{2} + 9 x = 4 x + 36

Move

36

to the left side

- x^{2} + 9 x - 36 = 4 x

Move

4 x

to the left side

- x^{2} + 5 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 1 ) ( - 36 )}{2 ( - 1 )}

Simplify

5^{2} - 4 (- 1) (- 36) : 119 i

x_{1, 2} = \frac{- 5 \pm 119 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 5 + 119 i}{2 ( - 1 )}, x_{2} = \frac{- 5 - 119 i}{2 ( - 1 )}

x = \frac{- 5 + 119 i}{2 ( - 1 )} : \frac{5}{2} - i \frac{119}{2}

x = \frac{- 5 - 119 i}{2 ( - 1 )} : \frac{5}{2} + i \frac{119}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} - i \frac{119}{2}, x = \frac{5}{2} + i \frac{119}{2}

3 + x - x^{2}

4 x^{2} - 8 x - 10 = 0

6 x^{2} - 10 x = 4

(3 x + 4)^{2} = 2

2 x - 3 x^{2} = 5

What are the solutions to the equation -x^2+9x=4x+36 ?
The solutions to the equation -x^2+9x=4x+36 are x= 5/2-i(sqrt(119))/2 ,x= 5/2+i(sqrt(119))/2
Find the zeros of -x^2+9x=4x+36
The zeros of -x^2+9x=4x+36 are x= 5/2-i(sqrt(119))/2 ,x= 5/2+i(sqrt(119))/2