What are the solutions to the equation 5x^2-4x=-1 ?

The solutions to the equation 5x^2-4x=-1 are x= 2/5+i 1/5 ,x= 2/5-i 1/5

Find the zeros of 5x^2-4x=-1

The zeros of 5x^2-4x=-1 are x= 2/5+i 1/5 ,x= 2/5-i 1/5

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5 x^{2} - 4 x = - 1

x = \frac{2}{5} + i \frac{1}{5}, x = \frac{2}{5} - i \frac{1}{5}

Solution steps

5 x^{2} - 4 x = - 1

Move

1

to the left side

5 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 i}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 2 i}{2 \cdot 5}

x = \frac{- ( - 4 ) + 2 i}{2 \cdot 5} : \frac{2}{5} + i \frac{1}{5}

x = \frac{- ( - 4 ) - 2 i}{2 \cdot 5} : \frac{2}{5} - i \frac{1}{5}

The solutions to the quadratic equation are:

x = \frac{2}{5} + i \frac{1}{5}, x = \frac{2}{5} - i \frac{1}{5}

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

x^{2} = - 2 x + 80

y^{2} - 10 y + 41 = 0

λ^{2} + 4 = 0

F = C D + C (D + C) + D (D + C)

What are the solutions to the equation 5x^2-4x=-1 ?
The solutions to the equation 5x^2-4x=-1 are x= 2/5+i 1/5 ,x= 2/5-i 1/5
Find the zeros of 5x^2-4x=-1
The zeros of 5x^2-4x=-1 are x= 2/5+i 1/5 ,x= 2/5-i 1/5