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

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

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

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

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x (2 x + 1) = 3

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

Decimal Notation

x = 1, x = - 1.5

Solution steps

x (2 x + 1) = 3

Expand

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

2 x^{2} + x = 3

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 3) = 5

x_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 \cdot 2}, x_{2} = \frac{- 1 - 5}{2 \cdot 2}

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

x = \frac{- 1 - 5}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

1156 = x^{2}

2 x^{2} + 6 x = 56

2 x^{2} + 19 x + 24 = 0

(2 x - 8) (x + 4) = 0

x, 5 x^{2} + 6 x y + 5 y^{2} - 4 x + 4 y = 0

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