Solve for y y^2+y=20

$y^{2} + y = 20$

Subtract $20$ from both sides of the equation.

$y^{2} + y - 20 = 0$

Factor $y^{2} + y - 20$ 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 $- 20$ and whose sum is $1$ .

$- 4, 5$

Write the factored form using these integers.

$(y - 4) (y + 5) = 0$

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

$y - 4 = 0$

$y + 5 = 0$

Set $y - 4$ equal to $0$ and solve for $y$ .

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

Set $y + 5$ equal to $0$ and solve for $y$ .

Set $y + 5$ equal to $0$ .

$y + 5 = 0$

Subtract $5$ from both sides of the equation.

$y = - 5$

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

$y = 4, - 5$

Solve for y y^2+y=20

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