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

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

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

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

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5 x^{2} + 4 x = 1

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

Decimal Notation

x = 0.2, x = - 1

Solution steps

5 x^{2} + 4 x = 1

Move

1

to the left side

5 x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 5 (- 1) = 6

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{2 + i}{3 + i}

- - 16

4 a^{2} + 6 = 0

3 x - 6 > \frac{1}{2}

2 x^{4} + 4 x^{3} - 30 x^{2}

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