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

The solutions to the equation x^2-5x+11=0 are x= 5/2+i(sqrt(19))/2 ,x= 5/2-i(sqrt(19))/2

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

The zeros of x^2-5x+11=0 are x= 5/2+i(sqrt(19))/2 ,x= 5/2-i(sqrt(19))/2

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

x = \frac{5}{2} + i \frac{19}{2}, x = \frac{5}{2} - i \frac{19}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 11 : 19 i

x_{1, 2} = \frac{- ( - 5 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 19 i}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 19 i}{2 \cdot 1}

x = \frac{- ( - 5 ) + 19 i}{2 \cdot 1} : \frac{5}{2} + i \frac{19}{2}

x = \frac{- ( - 5 ) - 19 i}{2 \cdot 1} : \frac{5}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{19}{2}, x = \frac{5}{2} - i \frac{19}{2}

x^{2} - 5 x = 3 x - 4 x

x^{2} - 2 x + 2 = 1

x, x^{2} y^{3} + 2 y = 6 x - 3

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

q, \frac{5 q ^{2}}{R} = \frac{B}{a}

What are the solutions to the equation x^2-5x+11=0 ?
The solutions to the equation x^2-5x+11=0 are x= 5/2+i(sqrt(19))/2 ,x= 5/2-i(sqrt(19))/2
Find the zeros of x^2-5x+11=0
The zeros of x^2-5x+11=0 are x= 5/2+i(sqrt(19))/2 ,x= 5/2-i(sqrt(19))/2