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

The solutions to the equation 5x^2+4x=2x^2-5x are x=0,x=-3

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

The zeros of 5x^2+4x=2x^2-5x are x=0,x=-3

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

x = 0, x = - 3

Solution steps

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

Move

5 x

to the left side

5 x^{2} + 9 x = 2 x^{2}

Move

2 x^{2}

to the left side

3 x^{2} + 9 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

9^{2} - 4 \cdot 3 \cdot 0 = 9

x_{1, 2} = \frac{- 9 \pm 9}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 9}{2 \cdot 3}, x_{2} = \frac{- 9 - 9}{2 \cdot 3}

x = \frac{- 9 + 9}{2 \cdot 3} : 0

x = \frac{- 9 - 9}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

x = 0, x = - 3

28 x^{2} - 85 x + 63 = 0

v^{2} = - 256

k, (n - k) (- k) = k (n - k)

4 x^{2} - 28 x + 57 = 0

- c^{2} + c + 1 = 0

What are the solutions to the equation 5x^2+4x=2x^2-5x ?
The solutions to the equation 5x^2+4x=2x^2-5x are x=0,x=-3
Find the zeros of 5x^2+4x=2x^2-5x
The zeros of 5x^2+4x=2x^2-5x are x=0,x=-3