Solve for d d^2+d-7=-5

$d^{2} + d - 7 = - 5$

Step 1

Add $5$ to both sides of the equation.

$d^{2} + d - 7 + 5 = 0$

Step 2

Add $- 7$ and $5$ .

$d^{2} + d - 2 = 0$

Step 3

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

$- 1, 2$

Write the factored form using these integers.

$(d - 1) (d + 2) = 0$

Step 4

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

$d - 1 = 0$

$d + 2 = 0$

Step 5

Set $d - 1$ equal to $0$ and solve for $d$ .

Set $d - 1$ equal to $0$ .

$d - 1 = 0$

Add $1$ to both sides of the equation.

$d = 1$

Step 6

Set $d + 2$ equal to $0$ and solve for $d$ .

Set $d + 2$ equal to $0$ .

$d + 2 = 0$

Subtract $2$ from both sides of the equation.

$d = - 2$

Step 7

The final solution is all the values that make $(d - 1) (d + 2) = 0$ true.

$d = 1, - 2$

Solve for d d^2+d-7=-5

Related Questions

Support LCMGCF.com by disabling your adblocker.