What are the solutions to the equation 7x^2-2x-3=0 ?

The solutions to the equation 7x^2-2x-3=0 are x=(1+sqrt(22))/7 ,x=(1-sqrt(22))/7

Find the zeros of 7x^2-2x-3=0

The zeros of 7x^2-2x-3=0 are x=(1+sqrt(22))/7 ,x=(1-sqrt(22))/7

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

x = \frac{1 + 22}{7}, x = \frac{1 - 22}{7}

Decimal Notation

x = 0.81291 \dots, x = - 0.52720 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 7 (- 3) = 222

x_{1, 2} = \frac{- ( - 2 ) \pm 2 22}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 22}{2 \cdot 7}, x_{2} = \frac{- ( - 2 ) - 2 22}{2 \cdot 7}

x = \frac{- ( - 2 ) + 2 22}{2 \cdot 7} : \frac{1 + 22}{7}

x = \frac{- ( - 2 ) - 2 22}{2 \cdot 7} : \frac{1 - 22}{7}

The solutions to the quadratic equation are:

x = \frac{1 + 22}{7}, x = \frac{1 - 22}{7}

1 - \frac{1}{9}

\frac{5}{4} x - \frac{2}{3} \geq \frac{8}{3} x - \frac{3}{2}

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

x^{2} - 6 x = - 1

- 4 (3 - x) = \frac{4}{5} (x + 5)

What are the solutions to the equation 7x^2-2x-3=0 ?
The solutions to the equation 7x^2-2x-3=0 are x=(1+sqrt(22))/7 ,x=(1-sqrt(22))/7
Find the zeros of 7x^2-2x-3=0
The zeros of 7x^2-2x-3=0 are x=(1+sqrt(22))/7 ,x=(1-sqrt(22))/7