What are the solutions to the equation 2x^2+12x-7=0 ?

The solutions to the equation 2x^2+12x-7=0 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2

Find the zeros of 2x^2+12x-7=0

The zeros of 2x^2+12x-7=0 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2

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2 x^{2} + 12 x - 7 = 0

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

Decimal Notation

x = 0.53553 \dots, x = - 6.53553 \dots

Solution steps

2 x^{2} + 12 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 7) = 102

x_{1, 2} = \frac{- 12 \pm 10 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 10 2}{2 \cdot 2}, x_{2} = \frac{- 12 - 10 2}{2 \cdot 2}

x = \frac{- 12 + 10 2}{2 \cdot 2} : \frac{- 6 + 5 2}{2}

x = \frac{- 12 - 10 2}{2 \cdot 2} : - \frac{6 + 5 2}{2}

The solutions to the quadratic equation are:

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

x^{2} (x - 4) \leq 0

x^{2} - x - 90

\frac{x}{4} - \frac{1}{10} \leq \frac{x}{5} + 1

125 r^{2} - 20

3 (1 - 2 x) = 6 x + 1

What are the solutions to the equation 2x^2+12x-7=0 ?
The solutions to the equation 2x^2+12x-7=0 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2
Find the zeros of 2x^2+12x-7=0
The zeros of 2x^2+12x-7=0 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2