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

The solutions to the equation 5x^2-3x+2=0 are x= 3/10+i(sqrt(31))/(10),x= 3/10-i(sqrt(31))/(10)

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

The zeros of 5x^2-3x+2=0 are x= 3/10+i(sqrt(31))/(10),x= 3/10-i(sqrt(31))/(10)

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

x = \frac{3}{10} + i \frac{31}{10}, x = \frac{3}{10} - i \frac{31}{10}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 5 \cdot 2 : 31 i

x_{1, 2} = \frac{- ( - 3 ) \pm 31 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 31 i}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 31 i}{2 \cdot 5}

x = \frac{- ( - 3 ) + 31 i}{2 \cdot 5} : \frac{3}{10} + i \frac{31}{10}

x = \frac{- ( - 3 ) - 31 i}{2 \cdot 5} : \frac{3}{10} - i \frac{31}{10}

The solutions to the quadratic equation are:

x = \frac{3}{10} + i \frac{31}{10}, x = \frac{3}{10} - i \frac{31}{10}

f = 2^{- 1}

2^{\frac{5}{4}} \cdot 3^{\frac{5}{4}}

4 k^{2} - 12 k + 9

3 x + 1 = x - 2

\frac{4 \cdot 1 0 ^{2019} - 1}{4 \cdot ( 3 \cdot 2019 ) + 1}

What are the solutions to the equation 5x^2-3x+2=0 ?
The solutions to the equation 5x^2-3x+2=0 are x= 3/10+i(sqrt(31))/(10),x= 3/10-i(sqrt(31))/(10)
Find the zeros of 5x^2-3x+2=0
The zeros of 5x^2-3x+2=0 are x= 3/10+i(sqrt(31))/(10),x= 3/10-i(sqrt(31))/(10)