What are the solutions to the equation (x+2)^2-(2x+7)/3 =3 ?

The solutions to the equation (x+2)^2-(2x+7)/3 =3 are x=(-5+sqrt(37))/3 ,x=-(5+sqrt(37))/3

Find the zeros of (x+2)^2-(2x+7)/3 =3

The zeros of (x+2)^2-(2x+7)/3 =3 are x=(-5+sqrt(37))/3 ,x=-(5+sqrt(37))/3

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(x + 2)^{2} - \frac{2 x + 7}{3} = 3

x = \frac{- 5 + 37}{3}, x = - \frac{5 + 37}{3}

Decimal Notation

x = 0.36092 \dots, x = - 3.69425 \dots

Solution steps

(x + 2)^{2} - \frac{2 x + 7}{3} = 3

Multiply both sides by

3

3 (x + 2)^{2} - (2 x + 7) = 9

Expand

3 (x + 2)^{2} - (2 x + 7) : 3 x^{2} + 10 x + 5

3 x^{2} + 10 x + 5 = 9

Move

9

to the left side

3 x^{2} + 10 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 (- 4) = 237

x_{1, 2} = \frac{- 10 \pm 2 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 10 + 2 37}{2 \cdot 3}, x_{2} = \frac{- 10 - 2 37}{2 \cdot 3}

x = \frac{- 10 + 2 37}{2 \cdot 3} : \frac{- 5 + 37}{3}

x = \frac{- 10 - 2 37}{2 \cdot 3} : - \frac{5 + 37}{3}

The solutions to the quadratic equation are:

x = \frac{- 5 + 37}{3}, x = - \frac{5 + 37}{3}

6 x^{2} + 3 x - 18 = 0

3 k^{2} - 18 k + 12 = 0

\frac{y ^{2} + 4}{8} = 5

- 5 (2 x - 10) (10 + 5 x) = 0

0 = - \frac{1}{4} x^{2} + 4 x - 1

What are the solutions to the equation (x+2)^2-(2x+7)/3 =3 ?
The solutions to the equation (x+2)^2-(2x+7)/3 =3 are x=(-5+sqrt(37))/3 ,x=-(5+sqrt(37))/3
Find the zeros of (x+2)^2-(2x+7)/3 =3
The zeros of (x+2)^2-(2x+7)/3 =3 are x=(-5+sqrt(37))/3 ,x=-(5+sqrt(37))/3