What are the solutions to the equation 9x^2-42x+50=0 ?

The solutions to the equation 9x^2-42x+50=0 are x= 7/3+i 1/3 ,x= 7/3-i 1/3

Find the zeros of 9x^2-42x+50=0

The zeros of 9x^2-42x+50=0 are x= 7/3+i 1/3 ,x= 7/3-i 1/3

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9 x^{2} - 42 x + 50 = 0

x = \frac{7}{3} + i \frac{1}{3}, x = \frac{7}{3} - i \frac{1}{3}

Solution steps

9 x^{2} - 42 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 9 \cdot 50}{2 \cdot 9}

Simplify

(- 42)^{2} - 4 \cdot 9 \cdot 50 : 6 i

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

Separate the solutions

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

x = \frac{- ( - 42 ) + 6 i}{2 \cdot 9} : \frac{7}{3} + i \frac{1}{3}

x = \frac{- ( - 42 ) - 6 i}{2 \cdot 9} : \frac{7}{3} - i \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{7}{3} + i \frac{1}{3}, x = \frac{7}{3} - i \frac{1}{3}

\frac{2}{3} (4)^{\frac{3}{2}}

\frac{98}{2}

- (\frac{2}{5})^{0}

3 2^{\frac{- 1}{5}}

- 8 (- 6 r + 19) + 194 = 32 r + 10

What are the solutions to the equation 9x^2-42x+50=0 ?
The solutions to the equation 9x^2-42x+50=0 are x= 7/3+i 1/3 ,x= 7/3-i 1/3
Find the zeros of 9x^2-42x+50=0
The zeros of 9x^2-42x+50=0 are x= 7/3+i 1/3 ,x= 7/3-i 1/3