What are the solutions to the equation 3x^2-6=(x+2)(x-3) ?

The solutions to the equation 3x^2-6=(x+2)(x-3) are x=-1/2 ,x=0

Find the zeros of 3x^2-6=(x+2)(x-3)

The zeros of 3x^2-6=(x+2)(x-3) are x=-1/2 ,x=0

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

x = - \frac{1}{2}, x = 0

Solution steps

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

Expand

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

3 x^{2} - 6 = x^{2} - x - 6

Switch sides

x^{2} - x - 6 = 3 x^{2} - 6

Move

6

to the left side

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

Move

3 x^{2}

to the left side

- 2 x^{2} - x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(- 1)^{2} - 4 (- 2) \cdot 0 = 1

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

Separate the solutions

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

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

x = \frac{- ( - 1 ) - 1}{2 ( - 2 )} : 0

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = 0

8 f^{2} + 11 f = 2 f^{2} + 3 (2 f - 5) + 21

y, 3 y - 2 x y - 3 y^{2} = 0

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

a x^{2} + 2 x + a = 5 x^{2} - 3 a x + 4

(36 - x)^{2} + (18 - x)^{2} = x^{2}

What are the solutions to the equation 3x^2-6=(x+2)(x-3) ?
The solutions to the equation 3x^2-6=(x+2)(x-3) are x=-1/2 ,x=0
Find the zeros of 3x^2-6=(x+2)(x-3)
The zeros of 3x^2-6=(x+2)(x-3) are x=-1/2 ,x=0