What are the solutions to the equation 0=x^2-3x+9 ?

The solutions to the equation 0=x^2-3x+9 are x= 3/2+i(3sqrt(3))/2 ,x= 3/2-i(3sqrt(3))/2

Find the zeros of 0=x^2-3x+9

The zeros of 0=x^2-3x+9 are x= 3/2+i(3sqrt(3))/2 ,x= 3/2-i(3sqrt(3))/2

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0 = x^{2} - 3 x + 9

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

Solution steps

0 = x^{2} - 3 x + 9

Switch sides

x^{2} - 3 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 9 : 33 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 3 t^{2} = - 27

(x)^{2} + (x + 1)^{2} + (x + 2)^{2} = 1454

m, F = \frac{G mm}{d ^{2}}

9 x^{2} + x = 0

c^{2} = 1 0^{2} + 6^{2}

What are the solutions to the equation 0=x^2-3x+9 ?
The solutions to the equation 0=x^2-3x+9 are x= 3/2+i(3sqrt(3))/2 ,x= 3/2-i(3sqrt(3))/2
Find the zeros of 0=x^2-3x+9
The zeros of 0=x^2-3x+9 are x= 3/2+i(3sqrt(3))/2 ,x= 3/2-i(3sqrt(3))/2