What are the solutions to the equation -x(x+7)=(x-2)(x+2) ?

The solutions to the equation -x(x+7)=(x-2)(x+2) are x=-4,x= 1/2

Find the zeros of -x(x+7)=(x-2)(x+2)

The zeros of -x(x+7)=(x-2)(x+2) are x=-4,x= 1/2

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- x (x + 7) = (x - 2) (x + 2)

x = - 4, x = \frac{1}{2}

Solution steps

- x (x + 7) = (x - 2) (x + 2)

Expand

- x (x + 7) : - x^{2} - 7 x

Expand

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

- x^{2} - 7 x = x^{2} - 4

Move

4

to the left side

- x^{2} - 7 x + 4 = x^{2}

Move

x^{2}

to the left side

- 2 x^{2} - 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

(- 7)^{2} - 4 (- 2) \cdot 4 = 9

x_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 9}{2 ( - 2 )}, x_{2} = \frac{- ( - 7 ) - 9}{2 ( - 2 )}

x = \frac{- ( - 7 ) + 9}{2 ( - 2 )} : - 4

x = \frac{- ( - 7 ) - 9}{2 ( - 2 )} : \frac{1}{2}

The solutions to the quadratic equation are:

x = - 4, x = \frac{1}{2}

x^{2} + 3 x + 4 = 0

2^{2^{2}}

3 a^{2} b^{2} + 18 ab

\frac{1 0 ^{- 14}}{1.8 \cdot 1 0 ^{- 5}}

\frac{( 1 + 1 i )}{( 1 + 1 i )}

What are the solutions to the equation -x(x+7)=(x-2)(x+2) ?
The solutions to the equation -x(x+7)=(x-2)(x+2) are x=-4,x= 1/2
Find the zeros of -x(x+7)=(x-2)(x+2)
The zeros of -x(x+7)=(x-2)(x+2) are x=-4,x= 1/2