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

The solutions to the equation 5x^2+4x=3 are x=(-2+sqrt(19))/5 ,x=-(2+sqrt(19))/5

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

The zeros of 5x^2+4x=3 are x=(-2+sqrt(19))/5 ,x=-(2+sqrt(19))/5

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5 x^{2} + 4 x = 3

x = \frac{- 2 + 19}{5}, x = - \frac{2 + 19}{5}

Solution steps

5 x^{2} + 4 x = 3

Move

3

to the left side

5 x^{2} + 4 x - 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 5 (- 3) = 219

x_{1, 2} = \frac{- 4 \pm 2 19}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 2 19}{2 \cdot 5}, x_{2} = \frac{- 4 - 2 19}{2 \cdot 5}

x = \frac{- 4 + 2 19}{2 \cdot 5} : \frac{- 2 + 19}{5}

x = \frac{- 4 - 2 19}{2 \cdot 5} : - \frac{2 + 19}{5}

The solutions to the quadratic equation are:

x = \frac{- 2 + 19}{5}, x = - \frac{2 + 19}{5}

5 x^{2} + 4 x = 4

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

What are the solutions to the equation 5x^2+4x=3 ?
The solutions to the equation 5x^2+4x=3 are x=(-2+sqrt(19))/5 ,x=-(2+sqrt(19))/5
Find the zeros of 5x^2+4x=3
The zeros of 5x^2+4x=3 are x=(-2+sqrt(19))/5 ,x=-(2+sqrt(19))/5