What are the solutions to the equation 9x^2-42x+52=0 ?

The solutions to the equation 9x^2-42x+52=0 are x= 7/3+i(sqrt(3))/3 ,x= 7/3-i(sqrt(3))/3

Find the zeros of 9x^2-42x+52=0

The zeros of 9x^2-42x+52=0 are x= 7/3+i(sqrt(3))/3 ,x= 7/3-i(sqrt(3))/3

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9 x^{2} - 42 x + 52 = 0

x = \frac{7}{3} + i \frac{3}{3}, x = \frac{7}{3} - i \frac{3}{3}

Solution steps

9 x^{2} - 42 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 9 \cdot 52}{2 \cdot 9}

Simplify

(- 42)^{2} - 4 \cdot 9 \cdot 52 : 63 i

x_{1, 2} = \frac{- ( - 42 ) \pm 6 3 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 6 3 i}{2 \cdot 9}, x_{2} = \frac{- ( - 42 ) - 6 3 i}{2 \cdot 9}

x = \frac{- ( - 42 ) + 6 3 i}{2 \cdot 9} : \frac{7}{3} + i \frac{3}{3}

x = \frac{- ( - 42 ) - 6 3 i}{2 \cdot 9} : \frac{7}{3} - i \frac{3}{3}

The solutions to the quadratic equation are:

x = \frac{7}{3} + i \frac{3}{3}, x = \frac{7}{3} - i \frac{3}{3}

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What are the solutions to the equation 9x^2-42x+52=0 ?
The solutions to the equation 9x^2-42x+52=0 are x= 7/3+i(sqrt(3))/3 ,x= 7/3-i(sqrt(3))/3
Find the zeros of 9x^2-42x+52=0
The zeros of 9x^2-42x+52=0 are x= 7/3+i(sqrt(3))/3 ,x= 7/3-i(sqrt(3))/3