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

The solutions to the equation 5x^2=-3x+3 are x=(-3+sqrt(69))/(10),x=(-3-sqrt(69))/(10)

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

The zeros of 5x^2=-3x+3 are x=(-3+sqrt(69))/(10),x=(-3-sqrt(69))/(10)

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

x = \frac{- 3 + 69}{10}, x = \frac{- 3 - 69}{10}

Solution steps

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

Move

3

to the left side

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

Move

3 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 5 (- 3) = 69

x_{1, 2} = \frac{- 3 \pm 69}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 3 + 69}{2 \cdot 5}, x_{2} = \frac{- 3 - 69}{2 \cdot 5}

x = \frac{- 3 + 69}{2 \cdot 5} : \frac{- 3 + 69}{10}

x = \frac{- 3 - 69}{2 \cdot 5} : \frac{- 3 - 69}{10}

The solutions to the quadratic equation are:

x = \frac{- 3 + 69}{10}, x = \frac{- 3 - 69}{10}

(\frac{4}{x})^{3}

\frac{2 π}{\frac{1}{3}}

\frac{1}{4} \cdot 6

\frac{x ^{8}}{x ^{2}}

2^{- \frac{1}{5}}

What are the solutions to the equation 5x^2=-3x+3 ?
The solutions to the equation 5x^2=-3x+3 are x=(-3+sqrt(69))/(10),x=(-3-sqrt(69))/(10)
Find the zeros of 5x^2=-3x+3
The zeros of 5x^2=-3x+3 are x=(-3+sqrt(69))/(10),x=(-3-sqrt(69))/(10)