Solve for y y^2=13y-42

$y^{2} = 13 y - 42$

Subtract $13 y$ from both sides of the equation.

$y^{2} - 13 y = - 42$

Add $42$ to both sides of the equation.

$y^{2} - 13 y + 42 = 0$

Factor $y^{2} - 13 y + 42$ 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 $42$ and whose sum is $- 13$ .

$- 7, - 6$

Write the factored form using these integers.

$(y - 7) (y - 6) = 0$

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

$y - 7 = 0$

$y - 6 = 0$

Set $y - 7$ equal to $0$ and solve for $y$ .

Set $y - 7$ equal to $0$ .

$y - 7 = 0$

Add $7$ to both sides of the equation.

$y = 7$

Set $y - 6$ equal to $0$ and solve for $y$ .

Set $y - 6$ equal to $0$ .

$y - 6 = 0$

Add $6$ to both sides of the equation.

$y = 6$

The final solution is all the values that make $(y - 7) (y - 6) = 0$ true.

$y = 7, 6$

Solve for y y^2=13y-42

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