What are the solutions to the equation 2x(x-4)+5x(1-x)=3(x+1) ?

The solutions to the equation 2x(x-4)+5x(1-x)=3(x+1) are x=-1

Find the zeros of 2x(x-4)+5x(1-x)=3(x+1)

The zeros of 2x(x-4)+5x(1-x)=3(x+1) are x=-1

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2 x (x - 4) + 5 x (1 - x) = 3 (x + 1)

x = - 1

Solution steps

2 x (x - 4) + 5 x (1 - x) = 3 (x + 1)

Expand

2 x (x - 4) + 5 x (1 - x) : - 3 x^{2} - 3 x

Expand

3 (x + 1) : 3 x + 3

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

Move

3

to the left side

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

Move

3 x

to the left side

- 3 x^{2} - 6 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 3 ) ( - 3 )}{2 ( - 3 )}

(- 6)^{2} - 4 (- 3) (- 3) = 0

x_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 ( - 3 )}

x = \frac{- ( - 6 )}{2 ( - 3 )}

\frac{- ( - 6 )}{2 ( - 3 )} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

∣ x - 4 ∣ \leq 7

4 x + 1 = x + 7

x^{4} - 81 \geq 0

25 x^{2} + 49 = 70 x

\frac{x + 1}{8} = 8

What are the solutions to the equation 2x(x-4)+5x(1-x)=3(x+1) ?
The solutions to the equation 2x(x-4)+5x(1-x)=3(x+1) are x=-1
Find the zeros of 2x(x-4)+5x(1-x)=3(x+1)
The zeros of 2x(x-4)+5x(1-x)=3(x+1) are x=-1