What are the solutions to the equation (x+2)(2x+1)-(x-1)^2=19 ?

The solutions to the equation (x+2)(2x+1)-(x-1)^2=19 are x=2,x=-9

Find the zeros of (x+2)(2x+1)-(x-1)^2=19

The zeros of (x+2)(2x+1)-(x-1)^2=19 are x=2,x=-9

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(x + 2) (2 x + 1) - (x - 1)^{2} = 19

x = 2, x = - 9

Solution steps

(x + 2) (2 x + 1) - (x - 1)^{2} = 19

Expand

(x + 2) (2 x + 1) - (x - 1)^{2} : x^{2} + 7 x + 1

x^{2} + 7 x + 1 = 19

Move

19

to the left side

x^{2} + 7 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

x_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 11}{2 \cdot 1}, x_{2} = \frac{- 7 - 11}{2 \cdot 1}

x = \frac{- 7 + 11}{2 \cdot 1} : 2

x = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 2, x = - 9

6 x^{2} + 2 x + 2 = 0

x^{2} + 8 x = - 20

x^{2} - 4 x - 12 = 0

5 x^{2} - 39 x - 4 = 4

8 x^{2} - 32 = 0

What are the solutions to the equation (x+2)(2x+1)-(x-1)^2=19 ?
The solutions to the equation (x+2)(2x+1)-(x-1)^2=19 are x=2,x=-9
Find the zeros of (x+2)(2x+1)-(x-1)^2=19
The zeros of (x+2)(2x+1)-(x-1)^2=19 are x=2,x=-9