What are the solutions to the equation 5x^2-4x+9=0 ?

The solutions to the equation 5x^2-4x+9=0 are x= 2/5+i(sqrt(41))/5 ,x= 2/5-i(sqrt(41))/5

Find the zeros of 5x^2-4x+9=0

The zeros of 5x^2-4x+9=0 are x= 2/5+i(sqrt(41))/5 ,x= 2/5-i(sqrt(41))/5

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5 x^{2} - 4 x + 9 = 0

x = \frac{2}{5} + i \frac{41}{5}, x = \frac{2}{5} - i \frac{41}{5}

Solution steps

5 x^{2} - 4 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 \cdot 9}{2 \cdot 5}

Simplify

(- 4)^{2} - 4 \cdot 5 \cdot 9 : 241 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 41 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 41 i}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 2 41 i}{2 \cdot 5}

x = \frac{- ( - 4 ) + 2 41 i}{2 \cdot 5} : \frac{2}{5} + i \frac{41}{5}

x = \frac{- ( - 4 ) - 2 41 i}{2 \cdot 5} : \frac{2}{5} - i \frac{41}{5}

The solutions to the quadratic equation are:

x = \frac{2}{5} + i \frac{41}{5}, x = \frac{2}{5} - i \frac{41}{5}

5 x^{2} + 20 x - 25 = 0

0 = 36 x^{2} - 24 x

6 x^{2} + 4 x = 0

z, (z + 3)^{3} + (z - m)^{3} = 2 (z^{3} + 6 m^{3})

3 w^{2} + 27 w = 0

What are the solutions to the equation 5x^2-4x+9=0 ?
The solutions to the equation 5x^2-4x+9=0 are x= 2/5+i(sqrt(41))/5 ,x= 2/5-i(sqrt(41))/5
Find the zeros of 5x^2-4x+9=0
The zeros of 5x^2-4x+9=0 are x= 2/5+i(sqrt(41))/5 ,x= 2/5-i(sqrt(41))/5