What are the solutions to the equation x^2-3x+6=0 ?

The solutions to the equation x^2-3x+6=0 are x= 3/2+i(sqrt(15))/2 ,x= 3/2-i(sqrt(15))/2

Find the zeros of x^2-3x+6=0

The zeros of x^2-3x+6=0 are x= 3/2+i(sqrt(15))/2 ,x= 3/2-i(sqrt(15))/2

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x^{2} - 3 x + 6 = 0

x = \frac{3}{2} + i \frac{15}{2}, x = \frac{3}{2} - i \frac{15}{2}

Solution steps

x^{2} - 3 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 6 : 15 i

x_{1, 2} = \frac{- ( - 3 ) \pm 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15 i}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 15 i}{2 \cdot 1}

x = \frac{- ( - 3 ) + 15 i}{2 \cdot 1} : \frac{3}{2} + i \frac{15}{2}

x = \frac{- ( - 3 ) - 15 i}{2 \cdot 1} : \frac{3}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{15}{2}, x = \frac{3}{2} - i \frac{15}{2}

(\frac{5}{2})^{- 3}

(8 a b^{2})^{- \frac{1}{2}} (8 a b^{2})^{\frac{1}{2}}

34 3^{- \frac{2}{3}}

4 > ∣ - 3 x - 5 ∣

m^{2} + m + 1 = 0

What are the solutions to the equation x^2-3x+6=0 ?
The solutions to the equation x^2-3x+6=0 are x= 3/2+i(sqrt(15))/2 ,x= 3/2-i(sqrt(15))/2
Find the zeros of x^2-3x+6=0
The zeros of x^2-3x+6=0 are x= 3/2+i(sqrt(15))/2 ,x= 3/2-i(sqrt(15))/2