What are the solutions to the equation 2(x+3)-x^2-3x=0 ?

The solutions to the equation 2(x+3)-x^2-3x=0 are x=-3,x=2

Find the zeros of 2(x+3)-x^2-3x=0

The zeros of 2(x+3)-x^2-3x=0 are x=-3,x=2

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

x = - 3, x = 2

Solution steps

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

Expand

2 (x + 3) - x^{2} - 3 x : - x^{2} - x + 6

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 6 = 5

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - 3

x = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

x = - 3, x = 2

12 x^{2} - 36 x = - 27

520 = 3 (x - 3) (2.1 x - 3)

25 a^{2} - 60 a + 36 = 18

x^{2} + 7 x - 20 = 0

(x + 8) (x + 8) = 0

What are the solutions to the equation 2(x+3)-x^2-3x=0 ?
The solutions to the equation 2(x+3)-x^2-3x=0 are x=-3,x=2
Find the zeros of 2(x+3)-x^2-3x=0
The zeros of 2(x+3)-x^2-3x=0 are x=-3,x=2