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

The solutions to the equation 5x^2+4x=4 are x=(2(sqrt(6)-1))/5 ,x=-(2(1+sqrt(6)))/5

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

The zeros of 5x^2+4x=4 are x=(2(sqrt(6)-1))/5 ,x=-(2(1+sqrt(6)))/5

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5 x^{2} + 4 x = 4

x = \frac{2 ( 6 - 1 )}{5}, x = - \frac{2 ( 1 + 6 )}{5}

Solution steps

5 x^{2} + 4 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 5 (- 4) = 46

x_{1, 2} = \frac{- 4 \pm 4 6}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 4 6}{2 \cdot 5}, x_{2} = \frac{- 4 - 4 6}{2 \cdot 5}

x = \frac{- 4 + 4 6}{2 \cdot 5} : \frac{2 ( 6 - 1 )}{5}

x = \frac{- 4 - 4 6}{2 \cdot 5} : - \frac{2 ( 1 + 6 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 6 - 1 )}{5}, x = - \frac{2 ( 1 + 6 )}{5}

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

x^{2} + x + 13 = 4 + 7 x

12 y^{2} = - 5 y^{2} + 2

(x - 4) (x - 6) = - 1

3 x^{2} + 9 x

What are the solutions to the equation 5x^2+4x=4 ?
The solutions to the equation 5x^2+4x=4 are x=(2(sqrt(6)-1))/5 ,x=-(2(1+sqrt(6)))/5
Find the zeros of 5x^2+4x=4
The zeros of 5x^2+4x=4 are x=(2(sqrt(6)-1))/5 ,x=-(2(1+sqrt(6)))/5