Solve for w 525=w^2+4w

$525 = w^{2} + 4 w$

Rewrite the equation as $w^{2} + 4 w = 525$ .

$w^{2} + 4 w = 525$

Subtract $525$ from both sides of the equation.

$w^{2} + 4 w - 525 = 0$

Factor $w^{2} + 4 w - 525$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 525$ and whose sum is $4$ .

$- 21, 25$

Write the factored form using these integers.

$(w - 21) (w + 25) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$w - 21 = 0$

$w + 25 = 0$

Set $w - 21$ equal to $0$ and solve for $w$ .

Set $w - 21$ equal to $0$ .

$w - 21 = 0$

Add $21$ to both sides of the equation.

$w = 21$

Set $w + 25$ equal to $0$ and solve for $w$ .

Set $w + 25$ equal to $0$ .

$w + 25 = 0$

Subtract $25$ from both sides of the equation.

$w = - 25$

The final solution is all the values that make $(w - 21) (w + 25) = 0$ true.

$w = 21, - 25$

Solve for w 525=w^2+4w

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