Solve for q 4q^2+3q=3q^2-4q+18

$4 q^{2} + 3 q = 3 q^{2} - 4 q + 18$

Step 1

Move all terms containing $q$ to the left side of the equation.

Subtract $3 q^{2}$ from both sides of the equation.

$4 q^{2} + 3 q - 3 q^{2} = - 4 q + 18$

Add $4 q$ to both sides of the equation.

$4 q^{2} + 3 q - 3 q^{2} + 4 q = 18$

Subtract $3 q^{2}$ from $4 q^{2}$ .

$q^{2} + 3 q + 4 q = 18$

Add $3 q$ and $4 q$ .

$q^{2} + 7 q = 18$

Step 2

Subtract $18$ from both sides of the equation.

$q^{2} + 7 q - 18 = 0$

Step 3

Factor $q^{2} + 7 q - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Write the factored form using these integers.

$(q - 2) (q + 9) = 0$

Step 4

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

$q - 2 = 0$

$q + 9 = 0$

Step 5

Set $q - 2$ equal to $0$ and solve for $q$ .

Set $q - 2$ equal to $0$ .

$q - 2 = 0$

Add $2$ to both sides of the equation.

$q = 2$

Step 6

Set $q + 9$ equal to $0$ and solve for $q$ .

Set $q + 9$ equal to $0$ .

$q + 9 = 0$

Subtract $9$ from both sides of the equation.

$q = - 9$

Step 7

The final solution is all the values that make $(q - 2) (q + 9) = 0$ true.

$q = 2, - 9$

Solve for q 4q^2+3q=3q^2-4q+18

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