What are the solutions to the equation x^2+2x=5(2x+2)-1 ?

The solutions to the equation x^2+2x=5(2x+2)-1 are x=9,x=-1

Find the zeros of x^2+2x=5(2x+2)-1

The zeros of x^2+2x=5(2x+2)-1 are x=9,x=-1

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x^{2} + 2 x = 5 (2 x + 2) - 1

x = 9, x = - 1

Solution steps

x^{2} + 2 x = 5 (2 x + 2) - 1

Expand

5 (2 x + 2) - 1 : 10 x + 9

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

Move

9

to the left side

x^{2} + 2 x - 9 = 10 x

Move

10 x

to the left side

x^{2} - 8 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 9) = 10

x_{1, 2} = \frac{- ( - 8 ) \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 10}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 10}{2 \cdot 1}

x = \frac{- ( - 8 ) + 10}{2 \cdot 1} : 9

x = \frac{- ( - 8 ) - 10}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 9, x = - 1

7 x + 3 = 13 + x

6 x^{2} + x - 12

8 x - 14 < 50

\frac{1}{2} (2 - 1)

(3 b^{- 2})^{2} (a^{2} b^{4})^{3}

What are the solutions to the equation x^2+2x=5(2x+2)-1 ?
The solutions to the equation x^2+2x=5(2x+2)-1 are x=9,x=-1
Find the zeros of x^2+2x=5(2x+2)-1
The zeros of x^2+2x=5(2x+2)-1 are x=9,x=-1