What are the solutions to the equation 5x^2+7x-4=-7x-1 ?

The solutions to the equation 5x^2+7x-4=-7x-1 are x= 1/5 ,x=-3

Find the zeros of 5x^2+7x-4=-7x-1

The zeros of 5x^2+7x-4=-7x-1 are x= 1/5 ,x=-3

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5 x^{2} + 7 x - 4 = - 7 x - 1

x = \frac{1}{5}, x = - 3

Solution steps

5 x^{2} + 7 x - 4 = - 7 x - 1

Move

1

to the left side

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

Move

7 x

to the left side

5 x^{2} + 14 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

1 4^{2} - 4 \cdot 5 (- 3) = 16

x_{1, 2} = \frac{- 14 \pm 16}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 14 + 16}{2 \cdot 5}, x_{2} = \frac{- 14 - 16}{2 \cdot 5}

x = \frac{- 14 + 16}{2 \cdot 5} : \frac{1}{5}

x = \frac{- 14 - 16}{2 \cdot 5} : - 3

The solutions to the quadratic equation are:

x = \frac{1}{5}, x = - 3

x (2 x - 4) = (x - 2) (x - 2)

3 x^{2} - 1 = 3 - 2 x^{2}

(3 x + 2) (2 x - 1) = 0

0.3 x^{2} + 0.4 x - 1 = 0

36 x^{2} - 4964 = 0

What are the solutions to the equation 5x^2+7x-4=-7x-1 ?
The solutions to the equation 5x^2+7x-4=-7x-1 are x= 1/5 ,x=-3
Find the zeros of 5x^2+7x-4=-7x-1
The zeros of 5x^2+7x-4=-7x-1 are x= 1/5 ,x=-3