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

The solutions to the equation 5x^2=2x-3 are x= 1/5+i(sqrt(14))/5 ,x= 1/5-i(sqrt(14))/5

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

The zeros of 5x^2=2x-3 are x= 1/5+i(sqrt(14))/5 ,x= 1/5-i(sqrt(14))/5

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5 x^{2} = 2 x - 3

x = \frac{1}{5} + i \frac{14}{5}, x = \frac{1}{5} - i \frac{14}{5}

Solution steps

5 x^{2} = 2 x - 3

Move

3

to the left side

5 x^{2} + 3 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 5 \cdot 3 : 214 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 14 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 14 i}{2 \cdot 5}, x_{2} = \frac{- ( - 2 ) - 2 14 i}{2 \cdot 5}

x = \frac{- ( - 2 ) + 2 14 i}{2 \cdot 5} : \frac{1}{5} + i \frac{14}{5}

x = \frac{- ( - 2 ) - 2 14 i}{2 \cdot 5} : \frac{1}{5} - i \frac{14}{5}

The solutions to the quadratic equation are:

x = \frac{1}{5} + i \frac{14}{5}, x = \frac{1}{5} - i \frac{14}{5}

\frac{1}{5} (5)

24 = - \frac{v}{3}

\frac{1}{x - 2 y} - \frac{x ^{2} + 4 y ^{2}}{x ^{3} - 8 y ^{3}}

What are the solutions to the equation 5x^2=2x-3 ?
The solutions to the equation 5x^2=2x-3 are x= 1/5+i(sqrt(14))/5 ,x= 1/5-i(sqrt(14))/5
Find the zeros of 5x^2=2x-3
The zeros of 5x^2=2x-3 are x= 1/5+i(sqrt(14))/5 ,x= 1/5-i(sqrt(14))/5