What are the solutions to the equation 9x+x(x-4)=12-19 ?

The solutions to the equation 9x+x(x-4)=12-19 are x=-5/2+i(sqrt(3))/2 ,x=-5/2-i(sqrt(3))/2

Find the zeros of 9x+x(x-4)=12-19

The zeros of 9x+x(x-4)=12-19 are x=-5/2+i(sqrt(3))/2 ,x=-5/2-i(sqrt(3))/2

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9 x + x (x - 4) = 12 - 19

x = - \frac{5}{2} + i \frac{3}{2}, x = - \frac{5}{2} - i \frac{3}{2}

Solution steps

9 x + x (x - 4) = 12 - 19

Expand

9 x + x (x - 4) : x^{2} + 5 x

x^{2} + 5 x = - 7

Move

7

to the left side

x^{2} + 5 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

5^{2} - 4 \cdot 1 \cdot 7 : 3 i

x_{1, 2} = \frac{- 5 \pm 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 3 i}{2 \cdot 1}, x_{2} = \frac{- 5 - 3 i}{2 \cdot 1}

x = \frac{- 5 + 3 i}{2 \cdot 1} : - \frac{5}{2} + i \frac{3}{2}

x = \frac{- 5 - 3 i}{2 \cdot 1} : - \frac{5}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i \frac{3}{2}, x = - \frac{5}{2} - i \frac{3}{2}

4 x^{2} + 25 x = 21

x^{2} + (- 1) x - (- 1 + 7) = 0

- x^{2} + 16 x - 48 = 0

8 x^{2} - 4 x - 3 = 0

- 6 x^{2} = 16 x + 8

What are the solutions to the equation 9x+x(x-4)=12-19 ?
The solutions to the equation 9x+x(x-4)=12-19 are x=-5/2+i(sqrt(3))/2 ,x=-5/2-i(sqrt(3))/2
Find the zeros of 9x+x(x-4)=12-19
The zeros of 9x+x(x-4)=12-19 are x=-5/2+i(sqrt(3))/2 ,x=-5/2-i(sqrt(3))/2