What are the solutions to the equation 11x^2-11x+8=4 ?

The solutions to the equation 11x^2-11x+8=4 are x= 1/2+i(sqrt(55))/(22),x= 1/2-i(sqrt(55))/(22)

Find the zeros of 11x^2-11x+8=4

The zeros of 11x^2-11x+8=4 are x= 1/2+i(sqrt(55))/(22),x= 1/2-i(sqrt(55))/(22)

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11 x^{2} - 11 x + 8 = 4

x = \frac{1}{2} + i \frac{55}{22}, x = \frac{1}{2} - i \frac{55}{22}

Solution steps

11 x^{2} - 11 x + 8 = 4

Move

4

to the left side

11 x^{2} - 11 x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 11)^{2} - 4 \cdot 11 \cdot 4 : 55 i

x_{1, 2} = \frac{- ( - 11 ) \pm 55 i}{2 \cdot 11}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 55 i}{2 \cdot 11}, x_{2} = \frac{- ( - 11 ) - 55 i}{2 \cdot 11}

x = \frac{- ( - 11 ) + 55 i}{2 \cdot 11} : \frac{1}{2} + i \frac{55}{22}

x = \frac{- ( - 11 ) - 55 i}{2 \cdot 11} : \frac{1}{2} - i \frac{55}{22}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{55}{22}, x = \frac{1}{2} - i \frac{55}{22}

t^{2} - 5 t

\frac{j}{4} > 2

- 3 - 10147

x - 5 = 9

- 4.9 t^{2} + 20 = 0

What are the solutions to the equation 11x^2-11x+8=4 ?
The solutions to the equation 11x^2-11x+8=4 are x= 1/2+i(sqrt(55))/(22),x= 1/2-i(sqrt(55))/(22)
Find the zeros of 11x^2-11x+8=4
The zeros of 11x^2-11x+8=4 are x= 1/2+i(sqrt(55))/(22),x= 1/2-i(sqrt(55))/(22)