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

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

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

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

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7 x^{2} + 2 x = 3

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

Decimal Notation

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

Solution steps

7 x^{2} + 2 x = 3

Move

3

to the left side

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}

x^{2} + 10 x = 3 x + 44

0 = x^{2} + 4 x + 1

x + x^{2} - 25 = 15 x + 47

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

x^{2} - 4 = 23

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