What are the solutions to the equation 5x^2-3x+1=0 ?

The solutions to the equation 5x^2-3x+1=0 are x= 3/10+i(sqrt(11))/(10),x= 3/10-i(sqrt(11))/(10)

Find the zeros of 5x^2-3x+1=0

The zeros of 5x^2-3x+1=0 are x= 3/10+i(sqrt(11))/(10),x= 3/10-i(sqrt(11))/(10)

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5 x^{2} - 3 x + 1 = 0

x = \frac{3}{10} + i \frac{11}{10}, x = \frac{3}{10} - i \frac{11}{10}

Solution steps

5 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

Simplify

(- 3)^{2} - 4 \cdot 5 \cdot 1 : 11 i

x_{1, 2} = \frac{- ( - 3 ) \pm 11 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 11 i}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 11 i}{2 \cdot 5}

x = \frac{- ( - 3 ) + 11 i}{2 \cdot 5} : \frac{3}{10} + i \frac{11}{10}

x = \frac{- ( - 3 ) - 11 i}{2 \cdot 5} : \frac{3}{10} - i \frac{11}{10}

The solutions to the quadratic equation are:

x = \frac{3}{10} + i \frac{11}{10}, x = \frac{3}{10} - i \frac{11}{10}

3^{2} + x^{2} = (2 x - 3)^{2}

2 + 2 x = - 2

3 x^{2} - 39 = 0

\frac{2 x}{2} + 4 - 4 x = 6 + 2 x

\frac{( x + 13 ) ( x - 15 )}{x + 11} \leq 0

What are the solutions to the equation 5x^2-3x+1=0 ?
The solutions to the equation 5x^2-3x+1=0 are x= 3/10+i(sqrt(11))/(10),x= 3/10-i(sqrt(11))/(10)
Find the zeros of 5x^2-3x+1=0
The zeros of 5x^2-3x+1=0 are x= 3/10+i(sqrt(11))/(10),x= 3/10-i(sqrt(11))/(10)