What are the solutions to the equation 3x^2-4x=-7 ?

The solutions to the equation 3x^2-4x=-7 are x= 2/3+i(sqrt(17))/3 ,x= 2/3-i(sqrt(17))/3

Find the zeros of 3x^2-4x=-7

The zeros of 3x^2-4x=-7 are x= 2/3+i(sqrt(17))/3 ,x= 2/3-i(sqrt(17))/3

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3 x^{2} - 4 x = - 7

x = \frac{2}{3} + i \frac{17}{3}, x = \frac{2}{3} - i \frac{17}{3}

Solution steps

3 x^{2} - 4 x = - 7

Move

7

to the left side

3 x^{2} - 4 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 7 : 217 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 17 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 17 i}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 17 i}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 17 i}{2 \cdot 3} : \frac{2}{3} + i \frac{17}{3}

x = \frac{- ( - 4 ) - 2 17 i}{2 \cdot 3} : \frac{2}{3} - i \frac{17}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{17}{3}, x = \frac{2}{3} - i \frac{17}{3}

2^{3}

7 - 4 x = 3

25 m^{100} - 121 n^{16}

b + 5 \leq 9 b - 4

3 y^{2} - 4 y

What are the solutions to the equation 3x^2-4x=-7 ?
The solutions to the equation 3x^2-4x=-7 are x= 2/3+i(sqrt(17))/3 ,x= 2/3-i(sqrt(17))/3
Find the zeros of 3x^2-4x=-7
The zeros of 3x^2-4x=-7 are x= 2/3+i(sqrt(17))/3 ,x= 2/3-i(sqrt(17))/3