What are the solutions to the equation 5x^2+12x-4=0 ?

The solutions to the equation 5x^2+12x-4=0 are x=(2(sqrt(14)-3))/5 ,x=-(2(3+sqrt(14)))/5

Find the zeros of 5x^2+12x-4=0

The zeros of 5x^2+12x-4=0 are x=(2(sqrt(14)-3))/5 ,x=-(2(3+sqrt(14)))/5

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5 x^{2} + 12 x - 4 = 0

x = \frac{2 ( 14 - 3 )}{5}, x = - \frac{2 ( 3 + 14 )}{5}

Decimal Notation

x = 0.29666 \dots, x = - 2.69666 \dots

Solution steps

5 x^{2} + 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

1 2^{2} - 4 \cdot 5 (- 4) = 414

x_{1, 2} = \frac{- 12 \pm 4 14}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 12 + 4 14}{2 \cdot 5}, x_{2} = \frac{- 12 - 4 14}{2 \cdot 5}

x = \frac{- 12 + 4 14}{2 \cdot 5} : \frac{2 ( 14 - 3 )}{5}

x = \frac{- 12 - 4 14}{2 \cdot 5} : - \frac{2 ( 3 + 14 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 14 - 3 )}{5}, x = - \frac{2 ( 3 + 14 )}{5}

- 4 x^{2} - 48 x - 208 = 0

x^{2} + 10 = 6 x

x^{2} + 10 = 25

x, x^{2} + (\frac{4 x}{3})^{2} = 225

0 = x^{2} - 12 x

What are the solutions to the equation 5x^2+12x-4=0 ?
The solutions to the equation 5x^2+12x-4=0 are x=(2(sqrt(14)-3))/5 ,x=-(2(3+sqrt(14)))/5
Find the zeros of 5x^2+12x-4=0
The zeros of 5x^2+12x-4=0 are x=(2(sqrt(14)-3))/5 ,x=-(2(3+sqrt(14)))/5