What are the solutions to the equation (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 ?

The solutions to the equation (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 are x=1,x=-5

Find the zeros of (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1

The zeros of (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 are x=1,x=-5

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(2 x + 1)^{2} - (3 x + 5) (2 x - 1) + 3 (x^{2} + 3 x - 4) = 2 x - 1

x = 1, x = - 5

Solution steps

(2 x + 1)^{2} - (3 x + 5) (2 x - 1) + 3 (x^{2} + 3 x - 4) = 2 x - 1

Expand

(2 x + 1)^{2} - (3 x + 5) (2 x - 1) + 3 (x^{2} + 3 x - 4) : x^{2} + 6 x - 6

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

Move

1

to the left side

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

Move

2 x

to the left side

x^{2} + 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 6}{2 \cdot 1}, x_{2} = \frac{- 4 - 6}{2 \cdot 1}

x = \frac{- 4 + 6}{2 \cdot 1} : 1

x = \frac{- 4 - 6}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

- x^{2} + 4 x - 6 = 0

9 y^{2} - 16 = 0

2 y^{2} - 32 = 0

x, x^{2} + y^{2} - z^{2} = 1

7 x^{2} + 6 x - 1 = - 1

What are the solutions to the equation (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 ?
The solutions to the equation (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 are x=1,x=-5
Find the zeros of (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1
The zeros of (2x+1)^2-(3x+5)(2x-1)+3(x^2+3x-4)=2x-1 are x=1,x=-5