What are the solutions to the equation -4=3x^2-3x ?

The solutions to the equation -4=3x^2-3x are x= 1/2+i(sqrt(39))/6 ,x= 1/2-i(sqrt(39))/6

Find the zeros of -4=3x^2-3x

The zeros of -4=3x^2-3x are x= 1/2+i(sqrt(39))/6 ,x= 1/2-i(sqrt(39))/6

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

x = \frac{1}{2} + i \frac{39}{6}, x = \frac{1}{2} - i \frac{39}{6}

Solution steps

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

Switch sides

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

Move

4

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

x = \frac{- ( - 3 ) + 39 i}{2 \cdot 3} : \frac{1}{2} + i \frac{39}{6}

x = \frac{- ( - 3 ) - 39 i}{2 \cdot 3} : \frac{1}{2} - i \frac{39}{6}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{39}{6}, x = \frac{1}{2} - i \frac{39}{6}

A = x^{2} - 6 x - 16

x^{2} - 10 x + 9 = 0

4 x^{2} + 20 x + 23 = 0

x^{2} + 18 x + 42 = 0

3 x^{2} + 10 x + 9 = 0

What are the solutions to the equation -4=3x^2-3x ?
The solutions to the equation -4=3x^2-3x are x= 1/2+i(sqrt(39))/6 ,x= 1/2-i(sqrt(39))/6
Find the zeros of -4=3x^2-3x
The zeros of -4=3x^2-3x are x= 1/2+i(sqrt(39))/6 ,x= 1/2-i(sqrt(39))/6