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

The solutions to the equation (x+3)*(x-4)+x*(x+2)=2-x^2 are x=2,x=-7/3

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

The zeros of (x+3)*(x-4)+x*(x+2)=2-x^2 are x=2,x=-7/3

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(x + 3) \cdot (x - 4) + x \cdot (x + 2) = 2 - x^{2}

x = 2, x = - \frac{7}{3}

Decimal Notation

x = 2, x = - 2.33333 \dots

Solution steps

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

Expand

(x + 3) (x - 4) + x (x + 2) : 2 x^{2} + x - 12

2 x^{2} + x - 12 = 2 - x^{2}

Move

x^{2}

to the left side

3 x^{2} + x - 12 = 2

Move

2

to the left side

3 x^{2} + x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 14 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 14) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 3}, x_{2} = \frac{- 1 - 13}{2 \cdot 3}

x = \frac{- 1 + 13}{2 \cdot 3} : 2

x = \frac{- 1 - 13}{2 \cdot 3} : - \frac{7}{3}

The solutions to the quadratic equation are:

x = 2, x = - \frac{7}{3}

∣ x - 5 ∣ < 2

x^{2} - 17 x + 16 = 0

1 \leq y \leq 3

2 y^{3} + 5 y^{2} - 6 y - 15

\frac{- 3}{4} m + \frac{1}{2} m - \frac{1}{2} m = \frac{- 7}{12}

What are the solutions to the equation (x+3)*(x-4)+x*(x+2)=2-x^2 ?
The solutions to the equation (x+3)*(x-4)+x*(x+2)=2-x^2 are x=2,x=-7/3
Find the zeros of (x+3)*(x-4)+x*(x+2)=2-x^2
The zeros of (x+3)*(x-4)+x*(x+2)=2-x^2 are x=2,x=-7/3