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

The solutions to the equation 2x^2-5x+9=0 are x= 5/4+i(sqrt(47))/4 ,x= 5/4-i(sqrt(47))/4

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

The zeros of 2x^2-5x+9=0 are x= 5/4+i(sqrt(47))/4 ,x= 5/4-i(sqrt(47))/4

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

x = \frac{5}{4} + i \frac{47}{4}, x = \frac{5}{4} - i \frac{47}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 9 : 47 i

x_{1, 2} = \frac{- ( - 5 ) \pm 47 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 47 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 47 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 47 i}{2 \cdot 2} : \frac{5}{4} + i \frac{47}{4}

x = \frac{- ( - 5 ) - 47 i}{2 \cdot 2} : \frac{5}{4} - i \frac{47}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{47}{4}, x = \frac{5}{4} - i \frac{47}{4}

3 < 7 \cdot 3 - 25

152 \cdot \frac{π}{180}

- 3 n + 4 < 5 n + 8

49 a^{2} z - 168 a yc + 294 a^{2} c - 28 a yz

3 (z + 2) + 4 z = 28 - 2 (z + 2)

What are the solutions to the equation 2x^2-5x+9=0 ?
The solutions to the equation 2x^2-5x+9=0 are x= 5/4+i(sqrt(47))/4 ,x= 5/4-i(sqrt(47))/4
Find the zeros of 2x^2-5x+9=0
The zeros of 2x^2-5x+9=0 are x= 5/4+i(sqrt(47))/4 ,x= 5/4-i(sqrt(47))/4