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

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

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

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

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3 x^{2} + 2 x = 7

x = \frac{- 1 + 22}{3}, x = - \frac{1 + 22}{3}

Decimal Notation

x = 1.23013 \dots, x = - 1.89680 \dots

Solution steps

3 x^{2} + 2 x = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 2 \pm 2 22}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 22}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 22}{2 \cdot 3}

x = \frac{- 2 + 2 22}{2 \cdot 3} : \frac{- 1 + 22}{3}

x = \frac{- 2 - 2 22}{2 \cdot 3} : - \frac{1 + 22}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 22}{3}, x = - \frac{1 + 22}{3}

9 x^{2} - 6 x + 4 = 0

\frac{6 !}{5 ! 1 !}

\frac{3}{3}

x^{2} + 10 x + 25 = 81

2 y^{\frac{1}{2}} \cdot y

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