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

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

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

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

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

x = - 3, x = 1

Solution steps

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

Expand

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

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

Move

6 x

to the left side

- 2 x^{2} - 4 x + 6 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 2) \cdot 6 = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 ( - 2 )}, x_{2} = \frac{- ( - 4 ) - 8}{2 ( - 2 )}

x = \frac{- ( - 4 ) + 8}{2 ( - 2 )} : - 3

x = \frac{- ( - 4 ) - 8}{2 ( - 2 )} : 1

The solutions to the quadratic equation are:

x = - 3, x = 1

5 x - 7 \leq 7 x - 11 \leq 5 x + 9

(\frac{4}{49})^{\frac{3}{2}}

0 = - 16 x^{2} + 20 x

x + 8 + ∣ 4 - x ∣ = 0

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

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