What are the solutions to the equation 2x^2+7x-6=0 ?

The solutions to the equation 2x^2+7x-6=0 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

Find the zeros of 2x^2+7x-6=0

The zeros of 2x^2+7x-6=0 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

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2 x^{2} + 7 x - 6 = 0

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

Solution steps

2 x^{2} + 7 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 6) = 97

x_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 97}{2 \cdot 2}, x_{2} = \frac{- 7 - 97}{2 \cdot 2}

x = \frac{- 7 + 97}{2 \cdot 2} : \frac{- 7 + 97}{4}

x = \frac{- 7 - 97}{2 \cdot 2} : \frac{- 7 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

(\frac{1}{x})^{3}

7 n - 4 = 31

3 x \geq 48

\frac{x + 7}{3} = \frac{- 8}{5}

p - 4 = - 9 + p

What are the solutions to the equation 2x^2+7x-6=0 ?
The solutions to the equation 2x^2+7x-6=0 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4
Find the zeros of 2x^2+7x-6=0
The zeros of 2x^2+7x-6=0 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4