What are the solutions to the equation -x^2+1+2x(x+3)=8 ?

The solutions to the equation -x^2+1+2x(x+3)=8 are x=1,x=-7

Find the zeros of -x^2+1+2x(x+3)=8

The zeros of -x^2+1+2x(x+3)=8 are x=1,x=-7

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- x^{2} + 1 + 2 x (x + 3) = 8

x = 1, x = - 7

Solution steps

- x^{2} + 1 + 2 x (x + 3) = 8

Expand

- x^{2} + 1 + 2 x (x + 3) : x^{2} + 6 x + 1

x^{2} + 6 x + 1 = 8

Move

8

to the left side

x^{2} + 6 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 7) = 8

x_{1, 2} = \frac{- 6 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 8}{2 \cdot 1}, x_{2} = \frac{- 6 - 8}{2 \cdot 1}

x = \frac{- 6 + 8}{2 \cdot 1} : 1

x = \frac{- 6 - 8}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 1, x = - 7

z^{2} - 8 z + 25 = 0

2 f (0) = x^{2} - 3

4 x^{2} - 20 x + 28 = 3

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

- 4 x^{2} - 7 x = 0

What are the solutions to the equation -x^2+1+2x(x+3)=8 ?
The solutions to the equation -x^2+1+2x(x+3)=8 are x=1,x=-7
Find the zeros of -x^2+1+2x(x+3)=8
The zeros of -x^2+1+2x(x+3)=8 are x=1,x=-7