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

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

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

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

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

x = 2, x = - 2

Solution steps

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

Expand

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

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

Move

4

to the right side

x^{2} + 3 x = - x^{2} + 3 x + 8

Move

3 x

to the left side

x^{2} = - x^{2} + 8

Move

x^{2}

to the left side

2 x^{2} = 8

Divide both sides by

2

x^{2} = 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

(w + 2) (2 w - 1) = (w - 2) (w + 5) + 15

4 - 2 x^{2} = 0

2 x^{2} + 11 = 9 x

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

x^{2} - 6 x - 7

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