What are the solutions to the equation (2x-5)(x+1)+12x=0 ?

The solutions to the equation (2x-5)(x+1)+12x=0 are x= 1/2 ,x=-5

Find the zeros of (2x-5)(x+1)+12x=0

The zeros of (2x-5)(x+1)+12x=0 are x= 1/2 ,x=-5

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(2 x - 5) (x + 1) + 12 x = 0

x = \frac{1}{2}, x = - 5

Decimal Notation

x = 0.5, x = - 5

Solution steps

(2 x - 5) (x + 1) + 12 x = 0

Expand

(2 x - 5) (x + 1) + 12 x : 2 x^{2} + 9 x - 5

2 x^{2} + 9 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

9^{2} - 4 \cdot 2 (- 5) = 11

x_{1, 2} = \frac{- 9 \pm 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 9 + 11}{2 \cdot 2}, x_{2} = \frac{- 9 - 11}{2 \cdot 2}

x = \frac{- 9 + 11}{2 \cdot 2} : \frac{1}{2}

x = \frac{- 9 - 11}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 5

x^{2} - 14 x + 9 = - 60

2 (x - 3) = x (13 - x)

3 x^{2} + 2 = 4 x^{2} + x - 2

2 x - x (x + 3) + 5 = 20

x^{2} = - 2 x + 24

What are the solutions to the equation (2x-5)(x+1)+12x=0 ?
The solutions to the equation (2x-5)(x+1)+12x=0 are x= 1/2 ,x=-5
Find the zeros of (2x-5)(x+1)+12x=0
The zeros of (2x-5)(x+1)+12x=0 are x= 1/2 ,x=-5