What are the solutions to the equation 5x^2+4x-6=0 ?

The solutions to the equation 5x^2+4x-6=0 are x=(-2+sqrt(34))/5 ,x=-(2+sqrt(34))/5

Find the zeros of 5x^2+4x-6=0

The zeros of 5x^2+4x-6=0 are x=(-2+sqrt(34))/5 ,x=-(2+sqrt(34))/5

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5 x^{2} + 4 x - 6 = 0

x = \frac{- 2 + 34}{5}, x = - \frac{2 + 34}{5}

Decimal Notation

x = 0.76619 \dots, x = - 1.56619 \dots

Solution steps

5 x^{2} + 4 x - 6 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 5 (- 6) = 234

x_{1, 2} = \frac{- 4 \pm 2 34}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 2 34}{2 \cdot 5}, x_{2} = \frac{- 4 - 2 34}{2 \cdot 5}

x = \frac{- 4 + 2 34}{2 \cdot 5} : \frac{- 2 + 34}{5}

x = \frac{- 4 - 2 34}{2 \cdot 5} : - \frac{2 + 34}{5}

The solutions to the quadratic equation are:

x = \frac{- 2 + 34}{5}, x = - \frac{2 + 34}{5}

\frac{1}{2} \cdot 3

x^{2} - 16 \leq 0

\frac{y}{1 - x ^{2} y ^{2}} d x + \frac{x}{1 - x ^{2} y ^{2}} d y

x^{2} - 3 x \leq 0

\frac{1}{2} + \frac{1}{4}

What are the solutions to the equation 5x^2+4x-6=0 ?
The solutions to the equation 5x^2+4x-6=0 are x=(-2+sqrt(34))/5 ,x=-(2+sqrt(34))/5
Find the zeros of 5x^2+4x-6=0
The zeros of 5x^2+4x-6=0 are x=(-2+sqrt(34))/5 ,x=-(2+sqrt(34))/5