What are the solutions to the equation 2(x-1)-(2x+3)(x+3)=6x-5 ?

The solutions to the equation 2(x-1)-(2x+3)(x+3)=6x-5 are x=-6,x=-1/2

Find the zeros of 2(x-1)-(2x+3)(x+3)=6x-5

The zeros of 2(x-1)-(2x+3)(x+3)=6x-5 are x=-6,x=-1/2

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2 (x - 1) - (2 x + 3) (x + 3) = 6 x - 5

x = - 6, x = - \frac{1}{2}

Decimal Notation

x = - 6, x = - 0.5

Solution steps

2 (x - 1) - (2 x + 3) (x + 3) = 6 x - 5

Expand

2 (x - 1) - (2 x + 3) (x + 3) : - 2 x^{2} - 7 x - 11

- 2 x^{2} - 7 x - 11 = 6 x - 5

Move

5

to the left side

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

Move

6 x

to the left side

- 2 x^{2} - 13 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 ( - 2 ) ( - 6 )}{2 ( - 2 )}

(- 13)^{2} - 4 (- 2) (- 6) = 11

x_{1, 2} = \frac{- ( - 13 ) \pm 11}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 11}{2 ( - 2 )}, x_{2} = \frac{- ( - 13 ) - 11}{2 ( - 2 )}

x = \frac{- ( - 13 ) + 11}{2 ( - 2 )} : - 6

x = \frac{- ( - 13 ) - 11}{2 ( - 2 )} : - \frac{1}{2}

The solutions to the quadratic equation are:

x = - 6, x = - \frac{1}{2}

2 x^{2} + 14 x = - 1

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

- 32 = 8 y^{2}

3 (x + 7)^{2} + 17 = 59

- 4.19 x^{2} + 2 = 0

What are the solutions to the equation 2(x-1)-(2x+3)(x+3)=6x-5 ?
The solutions to the equation 2(x-1)-(2x+3)(x+3)=6x-5 are x=-6,x=-1/2
Find the zeros of 2(x-1)-(2x+3)(x+3)=6x-5
The zeros of 2(x-1)-(2x+3)(x+3)=6x-5 are x=-6,x=-1/2