What are the solutions to the equation (w+2)(2w-1)=(w-2)(w+5)+15 ?

The solutions to the equation (w+2)(2w-1)=(w-2)(w+5)+15 are w=sqrt(7),w=-sqrt(7)

Find the zeros of (w+2)(2w-1)=(w-2)(w+5)+15

The zeros of (w+2)(2w-1)=(w-2)(w+5)+15 are w=sqrt(7),w=-sqrt(7)

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

w = 7, w = - 7

Decimal Notation

w = 2.64575 \dots, w = - 2.64575 \dots

Solution steps

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

Expand

(w + 2) (2 w - 1) : 2 w^{2} + 3 w - 2

Expand

(w - 2) (w + 5) + 15 : w^{2} + 3 w + 5

2 w^{2} + 3 w - 2 = w^{2} + 3 w + 5

Move

2

to the right side

2 w^{2} + 3 w = w^{2} + 3 w + 7

Move

3 w

to the left side

2 w^{2} = w^{2} + 7

Move

w^{2}

to the left side

w^{2} = 7

For

x^{2} = f (a)

the solutions are

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

w = 7, w = - 7

4 - 2 x^{2} = 0

2 x^{2} + 11 = 9 x

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

x^{2} - 6 x - 7

x^{2} - 32 x + 4 = 0

What are the solutions to the equation (w+2)(2w-1)=(w-2)(w+5)+15 ?
The solutions to the equation (w+2)(2w-1)=(w-2)(w+5)+15 are w=sqrt(7),w=-sqrt(7)
Find the zeros of (w+2)(2w-1)=(w-2)(w+5)+15
The zeros of (w+2)(2w-1)=(w-2)(w+5)+15 are w=sqrt(7),w=-sqrt(7)