What are the solutions to the equation (2x+1)(x+1)+(2x-1)x=7 ?

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

Find the zeros of (2x+1)(x+1)+(2x-1)x=7

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

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(2 x + 1) (x + 1) + (2 x - 1) x = 7

x = 1, x = - \frac{3}{2}

Decimal Notation

x = 1, x = - 1.5

Solution steps

(2 x + 1) (x + 1) + (2 x - 1) x = 7

Expand

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

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

Move

7

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 6) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 4}, x_{2} = \frac{- 2 - 10}{2 \cdot 4}

x = \frac{- 2 + 10}{2 \cdot 4} : 1

x = \frac{- 2 - 10}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 1, x = - \frac{3}{2}

n (4 + 2 n) - 4 (n - 6) = 74

3 x^{2} - 22 x + 7 = 0

x, 4 x^{2} + 22 x = 12

2 x^{2} - 7 = 43

3 (x^{2} - 6) - 2 = 1

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