What are the solutions to the equation x^2-3x+36=0 ?

The solutions to the equation x^2-3x+36=0 are x= 3/2+i(3sqrt(15))/2 ,x= 3/2-i(3sqrt(15))/2

Find the zeros of x^2-3x+36=0

The zeros of x^2-3x+36=0 are x= 3/2+i(3sqrt(15))/2 ,x= 3/2-i(3sqrt(15))/2

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x^{2} - 3 x + 36 = 0

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

Solution steps

x^{2} - 3 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 36 : 315 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} - 6 x + 10

\frac{1}{2} + \frac{5}{\frac{1}{2}}

- 4 (4 - 8 a) = - 112

2 x - 1 \geq - 5

- 3 [- 5 - (9 + 2 x)] = 2 (3 x - 1)

What are the solutions to the equation x^2-3x+36=0 ?
The solutions to the equation x^2-3x+36=0 are x= 3/2+i(3sqrt(15))/2 ,x= 3/2-i(3sqrt(15))/2
Find the zeros of x^2-3x+36=0
The zeros of x^2-3x+36=0 are x= 3/2+i(3sqrt(15))/2 ,x= 3/2-i(3sqrt(15))/2