What are the solutions to the equation 3(x+2)^2-2(x-1)(x+1)=11x+18 ?

The solutions to the equation 3(x+2)^2-2(x-1)(x+1)=11x+18 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

Find the zeros of 3(x+2)^2-2(x-1)(x+1)=11x+18

The zeros of 3(x+2)^2-2(x-1)(x+1)=11x+18 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

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3 (x + 2)^{2} - 2 (x - 1) (x + 1) = 11 x + 18

x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

Decimal Notation

x = 1.56155 \dots, x = - 2.56155 \dots

Solution steps

3 (x + 2)^{2} - 2 (x - 1) (x + 1) = 11 x + 18

Expand

3 (x + 2)^{2} - 2 (x - 1) (x + 1) : x^{2} + 12 x + 14

x^{2} + 12 x + 14 = 11 x + 18

Move

18

to the left side

x^{2} + 12 x - 4 = 11 x

Move

11 x

to the left side

x^{2} + x - 4 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 4) = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 1}, x_{2} = \frac{- 1 - 17}{2 \cdot 1}

x = \frac{- 1 + 17}{2 \cdot 1} : \frac{- 1 + 17}{2}

x = \frac{- 1 - 17}{2 \cdot 1} : \frac{- 1 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

2 x^{2} - 8 x - 5 = 0

x, 4 x^{2} - 21 x - 18 = 0

6 x^{2} - 6 x = 12

5 x^{2} = - 6 x - 1

352 = 7 x^{2} + 9

What are the solutions to the equation 3(x+2)^2-2(x-1)(x+1)=11x+18 ?
The solutions to the equation 3(x+2)^2-2(x-1)(x+1)=11x+18 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2
Find the zeros of 3(x+2)^2-2(x-1)(x+1)=11x+18
The zeros of 3(x+2)^2-2(x-1)(x+1)=11x+18 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2