What are the solutions to the equation 3x^2+5x-4=x^2-2x ?

The solutions to the equation 3x^2+5x-4=x^2-2x are x= 1/2 ,x=-4

Find the zeros of 3x^2+5x-4=x^2-2x

The zeros of 3x^2+5x-4=x^2-2x are x= 1/2 ,x=-4

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

x = \frac{1}{2}, x = - 4

Decimal Notation

x = 0.5, x = - 4

Solution steps

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

Move

2 x

to the left side

3 x^{2} + 7 x - 4 = x^{2}

Move

x^{2}

to the left side

2 x^{2} + 7 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 4) = 9

x_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 9}{2 \cdot 2}, x_{2} = \frac{- 7 - 9}{2 \cdot 2}

x = \frac{- 7 + 9}{2 \cdot 2} : \frac{1}{2}

x = \frac{- 7 - 9}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 4

x^{2} = 16 x - 56

a^{2} + 3 a - 4 = 0

x^{2} + 7 x + 6 = 0

0 = \frac{12 - x ^{2}}{5}

- 6 x^{2} + 36 x = - 714

What are the solutions to the equation 3x^2+5x-4=x^2-2x ?
The solutions to the equation 3x^2+5x-4=x^2-2x are x= 1/2 ,x=-4
Find the zeros of 3x^2+5x-4=x^2-2x
The zeros of 3x^2+5x-4=x^2-2x are x= 1/2 ,x=-4