Solve for w 4w^2+4w+1=121

$4 w^{2} + 4 w + 1 = 121$

Subtract $121$ from both sides of the equation.

$4 w^{2} + 4 w + 1 - 121 = 0$

Subtract $121$ from $1$ .

$4 w^{2} + 4 w - 120 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 w^{2} + 4 w - 120$ .

Factor $4$ out of $4 w^{2}$ .

$4 (w^{2}) + 4 w - 120 = 0$

Factor $4$ out of $4 w$ .

$4 (w^{2}) + 4 (w) - 120 = 0$

Factor $4$ out of $- 120$ .

$4 (w^{2}) + 4 w + 4 \cdot - 30 = 0$

Factor $4$ out of $4 (w^{2}) + 4 w$ .

$4 (w^{2} + w) + 4 \cdot - 30 = 0$

Factor $4$ out of $4 (w^{2} + w) + 4 \cdot - 30$ .

$4 (w^{2} + w - 30) = 0$

Factor.

Factor $w^{2} + w - 30$ 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 $- 30$ and whose sum is $1$ .

$- 5, 6$

Write the factored form using these integers.

$4 ((w - 5) (w + 6)) = 0$

Remove unnecessary parentheses.

$4 (w - 5) (w + 6) = 0$

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

$w - 5 = 0$

$w + 6 = 0$

Set $w - 5$ equal to $0$ and solve for $w$ .

Set $w - 5$ equal to $0$ .

$w - 5 = 0$

Add $5$ to both sides of the equation.

$w = 5$

Set $w + 6$ equal to $0$ and solve for $w$ .

Set $w + 6$ equal to $0$ .

$w + 6 = 0$

Subtract $6$ from both sides of the equation.

$w = - 6$

The final solution is all the values that make $4 (w - 5) (w + 6) = 0$ true.

$w = 5, - 6$

Solve for w 4w^2+4w+1=121

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