What are the solutions to the equation 7x^2-3x=2 ?

The solutions to the equation 7x^2-3x=2 are x=(3+sqrt(65))/(14),x=(3-sqrt(65))/(14)

Find the zeros of 7x^2-3x=2

The zeros of 7x^2-3x=2 are x=(3+sqrt(65))/(14),x=(3-sqrt(65))/(14)

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7 x^{2} - 3 x = 2

x = \frac{3 + 65}{14}, x = \frac{3 - 65}{14}

Solution steps

7 x^{2} - 3 x = 2

Move

2

to the left side

7 x^{2} - 3 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 7 ( - 2 )}{2 \cdot 7}

(- 3)^{2} - 4 \cdot 7 (- 2) = 65

x_{1, 2} = \frac{- ( - 3 ) \pm 65}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 65}{2 \cdot 7}, x_{2} = \frac{- ( - 3 ) - 65}{2 \cdot 7}

x = \frac{- ( - 3 ) + 65}{2 \cdot 7} : \frac{3 + 65}{14}

x = \frac{- ( - 3 ) - 65}{2 \cdot 7} : \frac{3 - 65}{14}

The solutions to the quadratic equation are:

x = \frac{3 + 65}{14}, x = \frac{3 - 65}{14}

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What are the solutions to the equation 7x^2-3x=2 ?
The solutions to the equation 7x^2-3x=2 are x=(3+sqrt(65))/(14),x=(3-sqrt(65))/(14)
Find the zeros of 7x^2-3x=2
The zeros of 7x^2-3x=2 are x=(3+sqrt(65))/(14),x=(3-sqrt(65))/(14)