What are the solutions to the equation 3x^2-7x+6=0 ?

The solutions to the equation 3x^2-7x+6=0 are x= 7/6+i(sqrt(23))/6 ,x= 7/6-i(sqrt(23))/6

Find the zeros of 3x^2-7x+6=0

The zeros of 3x^2-7x+6=0 are x= 7/6+i(sqrt(23))/6 ,x= 7/6-i(sqrt(23))/6

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3 x^{2} - 7 x + 6 = 0

x = \frac{7}{6} + i \frac{23}{6}, x = \frac{7}{6} - i \frac{23}{6}

Solution steps

3 x^{2} - 7 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

Simplify

(- 7)^{2} - 4 \cdot 3 \cdot 6 : 23 i

x_{1, 2} = \frac{- ( - 7 ) \pm 23 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 23 i}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 23 i}{2 \cdot 3}

x = \frac{- ( - 7 ) + 23 i}{2 \cdot 3} : \frac{7}{6} + i \frac{23}{6}

x = \frac{- ( - 7 ) - 23 i}{2 \cdot 3} : \frac{7}{6} - i \frac{23}{6}

The solutions to the quadratic equation are:

x = \frac{7}{6} + i \frac{23}{6}, x = \frac{7}{6} - i \frac{23}{6}

9 x^{4} - y^{8}

\frac{2}{3} x - \frac{1}{2} \leq 0

- \frac{2}{3}

9 x^{2} + 4 = 0

\frac{( p ) ( a )}{2}

What are the solutions to the equation 3x^2-7x+6=0 ?
The solutions to the equation 3x^2-7x+6=0 are x= 7/6+i(sqrt(23))/6 ,x= 7/6-i(sqrt(23))/6
Find the zeros of 3x^2-7x+6=0
The zeros of 3x^2-7x+6=0 are x= 7/6+i(sqrt(23))/6 ,x= 7/6-i(sqrt(23))/6