What are the solutions to the equation 3x^2-7x+8=0 ?

The solutions to the equation 3x^2-7x+8=0 are x= 7/6+i(sqrt(47))/6 ,x= 7/6-i(sqrt(47))/6

Find the zeros of 3x^2-7x+8=0

The zeros of 3x^2-7x+8=0 are x= 7/6+i(sqrt(47))/6 ,x= 7/6-i(sqrt(47))/6

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3 x^{2} - 7 x + 8 = 0

x = \frac{7}{6} + i \frac{47}{6}, x = \frac{7}{6} - i \frac{47}{6}

Solution steps

3 x^{2} - 7 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

(- 7)^{2} - 4 \cdot 3 \cdot 8 : 47 i

x_{1, 2} = \frac{- ( - 7 ) \pm 47 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 47 i}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 47 i}{2 \cdot 3}

x = \frac{- ( - 7 ) + 47 i}{2 \cdot 3} : \frac{7}{6} + i \frac{47}{6}

x = \frac{- ( - 7 ) - 47 i}{2 \cdot 3} : \frac{7}{6} - i \frac{47}{6}

The solutions to the quadratic equation are:

x = \frac{7}{6} + i \frac{47}{6}, x = \frac{7}{6} - i \frac{47}{6}

(6 x^{4} y^{6})^{3}

- 2 x + 1 = x + 4

4 x^{2} + 21 x - 18

16 = \frac{k}{11}

\frac{2}{8}

What are the solutions to the equation 3x^2-7x+8=0 ?
The solutions to the equation 3x^2-7x+8=0 are x= 7/6+i(sqrt(47))/6 ,x= 7/6-i(sqrt(47))/6
Find the zeros of 3x^2-7x+8=0
The zeros of 3x^2-7x+8=0 are x= 7/6+i(sqrt(47))/6 ,x= 7/6-i(sqrt(47))/6