What are the solutions to the equation (3-x)(1+2x)=-9 ?

The solutions to the equation (3-x)(1+2x)=-9 are x=-3/2 ,x=4

Find the zeros of (3-x)(1+2x)=-9

The zeros of (3-x)(1+2x)=-9 are x=-3/2 ,x=4

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(3 - x) (1 + 2 x) = - 9

x = - \frac{3}{2}, x = 4

Solution steps

(3 - x) (1 + 2 x) = - 9

Expand

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

3 + 5 x - 2 x^{2} = - 9

Move

9

to the left side

- 2 x^{2} + 5 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) \cdot 12}{2 ( - 2 )}

5^{2} - 4 (- 2) \cdot 12 = 11

x_{1, 2} = \frac{- 5 \pm 11}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 5 + 11}{2 ( - 2 )}, x_{2} = \frac{- 5 - 11}{2 ( - 2 )}

x = \frac{- 5 + 11}{2 ( - 2 )} : - \frac{3}{2}

x = \frac{- 5 - 11}{2 ( - 2 )} : 4

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = 4

(x + 3)^{2} + (x + 2)^{2} = 9

4 = x^{2} + 10 x

2 x - 5 (x + 3) = x (x - 4)

3 x + 4 = 2 x^{2} - 7

407 x^{2} + 480 x - 69 = 0

What are the solutions to the equation (3-x)(1+2x)=-9 ?
The solutions to the equation (3-x)(1+2x)=-9 are x=-3/2 ,x=4
Find the zeros of (3-x)(1+2x)=-9
The zeros of (3-x)(1+2x)=-9 are x=-3/2 ,x=4