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

The solutions to the equation 5x^2-5x+4=0 are x= 1/2+i(sqrt(55))/(10),x= 1/2-i(sqrt(55))/(10)

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

The zeros of 5x^2-5x+4=0 are x= 1/2+i(sqrt(55))/(10),x= 1/2-i(sqrt(55))/(10)

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

x = \frac{1}{2} + i \frac{55}{10}, x = \frac{1}{2} - i \frac{55}{10}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 5 \cdot 4 : 55 i

x_{1, 2} = \frac{- ( - 5 ) \pm 55 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 55 i}{2 \cdot 5}, x_{2} = \frac{- ( - 5 ) - 55 i}{2 \cdot 5}

x = \frac{- ( - 5 ) + 55 i}{2 \cdot 5} : \frac{1}{2} + i \frac{55}{10}

x = \frac{- ( - 5 ) - 55 i}{2 \cdot 5} : \frac{1}{2} - i \frac{55}{10}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{55}{10}, x = \frac{1}{2} - i \frac{55}{10}

y^{2} - 4 y - 60

(\frac{1}{5})^{- 5}

57

\frac{7}{8} \cdot 40

r^{2} + 16 r + 64 - q^{2}

What are the solutions to the equation 5x^2-5x+4=0 ?
The solutions to the equation 5x^2-5x+4=0 are x= 1/2+i(sqrt(55))/(10),x= 1/2-i(sqrt(55))/(10)
Find the zeros of 5x^2-5x+4=0
The zeros of 5x^2-5x+4=0 are x= 1/2+i(sqrt(55))/(10),x= 1/2-i(sqrt(55))/(10)