What are the solutions to the equation 2x^2-5x+11=0 ?

The solutions to the equation 2x^2-5x+11=0 are x= 5/4+i(3sqrt(7))/4 ,x= 5/4-i(3sqrt(7))/4

Find the zeros of 2x^2-5x+11=0

The zeros of 2x^2-5x+11=0 are x= 5/4+i(3sqrt(7))/4 ,x= 5/4-i(3sqrt(7))/4

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2 x^{2} - 5 x + 11 = 0

x = \frac{5}{4} + i \frac{3 7}{4}, x = \frac{5}{4} - i \frac{3 7}{4}

Solution steps

2 x^{2} - 5 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 11}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 11 : 37 i

x_{1, 2} = \frac{- ( - 5 ) \pm 3 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 3 7 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 3 7 i}{2 \cdot 2} : \frac{5}{4} + i \frac{3 7}{4}

x = \frac{- ( - 5 ) - 3 7 i}{2 \cdot 2} : \frac{5}{4} - i \frac{3 7}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{3 7}{4}, x = \frac{5}{4} - i \frac{3 7}{4}

y + 6 < 10 or 2 y - 3 > 9

- 2 x^{4} + 13 x^{3} - 21 x^{2} + 2 x + 8

- 3 x + 2 = 0

(\frac{1}{25})^{2}

\frac{3}{8} + \frac{3}{2}

What are the solutions to the equation 2x^2-5x+11=0 ?
The solutions to the equation 2x^2-5x+11=0 are x= 5/4+i(3sqrt(7))/4 ,x= 5/4-i(3sqrt(7))/4
Find the zeros of 2x^2-5x+11=0
The zeros of 2x^2-5x+11=0 are x= 5/4+i(3sqrt(7))/4 ,x= 5/4-i(3sqrt(7))/4