What are the solutions to the equation 3x^2-4x=-8 ?

The solutions to the equation 3x^2-4x=-8 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3

Find the zeros of 3x^2-4x=-8

The zeros of 3x^2-4x=-8 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3

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3 x^{2} - 4 x = - 8

x = \frac{2}{3} + i \frac{2 5}{3}, x = \frac{2}{3} - i \frac{2 5}{3}

Solution steps

3 x^{2} - 4 x = - 8

Move

8

to the left side

3 x^{2} - 4 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 8 : 45 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 5 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 5 i}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 4 5 i}{2 \cdot 3}

x = \frac{- ( - 4 ) + 4 5 i}{2 \cdot 3} : \frac{2}{3} + i \frac{2 5}{3}

x = \frac{- ( - 4 ) - 4 5 i}{2 \cdot 3} : \frac{2}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{2 5}{3}, x = \frac{2}{3} - i \frac{2 5}{3}

- \frac{7}{8} + \frac{1}{4}

4 x - 2 x^{2} \geq 0

25 + 36

3.5 n - 1356 - 1.28 \cdot 2.916 n > 0

4 x^{2} + 8 x - 10 = 0

What are the solutions to the equation 3x^2-4x=-8 ?
The solutions to the equation 3x^2-4x=-8 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3
Find the zeros of 3x^2-4x=-8
The zeros of 3x^2-4x=-8 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3