What are the solutions to the equation 9x^2-42x+53=0 ?

The solutions to the equation 9x^2-42x+53=0 are x= 7/3+i 2/3 ,x= 7/3-i 2/3

Find the zeros of 9x^2-42x+53=0

The zeros of 9x^2-42x+53=0 are x= 7/3+i 2/3 ,x= 7/3-i 2/3

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9 x^{2} - 42 x + 53 = 0

x = \frac{7}{3} + i \frac{2}{3}, x = \frac{7}{3} - i \frac{2}{3}

Solution steps

9 x^{2} - 42 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 9 \cdot 53}{2 \cdot 9}

Simplify

(- 42)^{2} - 4 \cdot 9 \cdot 53 : 12 i

x_{1, 2} = \frac{- ( - 42 ) \pm 12 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 12 i}{2 \cdot 9}, x_{2} = \frac{- ( - 42 ) - 12 i}{2 \cdot 9}

x = \frac{- ( - 42 ) + 12 i}{2 \cdot 9} : \frac{7}{3} + i \frac{2}{3}

x = \frac{- ( - 42 ) - 12 i}{2 \cdot 9} : \frac{7}{3} - i \frac{2}{3}

The solutions to the quadratic equation are:

x = \frac{7}{3} + i \frac{2}{3}, x = \frac{7}{3} - i \frac{2}{3}

h^{2} - 60 h + 100 = 0

5 (x - 5)^{2} - 33 = - 28

4 (x - 4)^{2} - 33 = - 5

2 v^{2} + 15 v + 19 = 0

0 = 2 x^{2} - 10 x + 7

What are the solutions to the equation 9x^2-42x+53=0 ?
The solutions to the equation 9x^2-42x+53=0 are x= 7/3+i 2/3 ,x= 7/3-i 2/3
Find the zeros of 9x^2-42x+53=0
The zeros of 9x^2-42x+53=0 are x= 7/3+i 2/3 ,x= 7/3-i 2/3