What are the solutions to the equation x^2-3x+12=0 ?

The solutions to the equation x^2-3x+12=0 are x= 3/2+i(sqrt(39))/2 ,x= 3/2-i(sqrt(39))/2

Find the zeros of x^2-3x+12=0

The zeros of x^2-3x+12=0 are x= 3/2+i(sqrt(39))/2 ,x= 3/2-i(sqrt(39))/2

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x^{2} - 3 x + 12 = 0

x = \frac{3}{2} + i \frac{39}{2}, x = \frac{3}{2} - i \frac{39}{2}

Solution steps

x^{2} - 3 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 12 : 39 i

x_{1, 2} = \frac{- ( - 3 ) \pm 39 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 39 i}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 39 i}{2 \cdot 1}

x = \frac{- ( - 3 ) + 39 i}{2 \cdot 1} : \frac{3}{2} + i \frac{39}{2}

x = \frac{- ( - 3 ) - 39 i}{2 \cdot 1} : \frac{3}{2} - i \frac{39}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{39}{2}, x = \frac{3}{2} - i \frac{39}{2}

\frac{7}{24.4} = \frac{x}{15.4841}

\frac{\frac{1}{( t + h ) ^{2}} - \frac{1}{t ^{2}}}{h}

- 7 < \frac{5 x - 1}{2} < 0

12 x^{2} + 14 x - 20

6^{2} + 24 (2) \leq \frac{5 x + 4 x}{2}

What are the solutions to the equation x^2-3x+12=0 ?
The solutions to the equation x^2-3x+12=0 are x= 3/2+i(sqrt(39))/2 ,x= 3/2-i(sqrt(39))/2
Find the zeros of x^2-3x+12=0
The zeros of x^2-3x+12=0 are x= 3/2+i(sqrt(39))/2 ,x= 3/2-i(sqrt(39))/2