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

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

Solve for q

$$q=-9$$

$$q=2$$

Steps Using Factoring

Subtract 3q^{2} from both sides.

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

Combine 4q^{2} and -3q^{2} to get q^{2}.

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

Add 4q to both sides.

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

Combine 3q and 4q to get 7q.

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

Subtract 18 from both sides.

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

To solve the equation, factor q^{2}+7q-18 using formula q^{2}+\left(a+b\right)q+ab=\left(q+a\right)\left(q+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=-2 b=9$$

Rewrite factored expression \left(q+a\right)\left(q+b\right) using the obtained values.

$$\left(q-2\right)\left(q+9\right)$$

To find equation solutions, solve q-2=0 and q+9=0.

$$q=2 q=-9$$

Steps Using Factoring By Grouping

Subtract 3q^{2} from both sides.

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

Combine 4q^{2} and -3q^{2} to get q^{2}.

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

Add 4q to both sides.

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

Combine 3q and 4q to get 7q.

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

Subtract 18 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as q^{2}+aq+bq-18. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-18\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=-2 b=9$$

Rewrite q^{2}+7q-18 as \left(q^{2}-2q\right)+\left(9q-18\right).

$$\left(q^{2}-2q\right)+\left(9q-18\right)$$

Factor out q in the first and 9 in the second group.

$$q\left(q-2\right)+9\left(q-2\right)$$

Factor out common term q-2 by using distributive property.

$$\left(q-2\right)\left(q+9\right)$$

To find equation solutions, solve q-2=0 and q+9=0.

$$q=2 q=-9$$

Steps Using the Quadratic Formula

Subtract 3q^{2} from both sides.

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

Combine 4q^{2} and -3q^{2} to get q^{2}.

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

Add 4q to both sides.

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

Combine 3q and 4q to get 7q.

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

Subtract 18 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-7±\sqrt{7^{2}-4\left(-18\right)}}{2}$$

Square 7.

$$q=\frac{-7±\sqrt{49-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$q=\frac{-7±\sqrt{49+72}}{2}$$

Add 49 to 72.

$$q=\frac{-7±\sqrt{121}}{2}$$

Take the square root of 121.

$$q=\frac{-7±11}{2}$$

Now solve the equation q=\frac{-7±11}{2} when ± is plus. Add -7 to 11.

$$q=\frac{4}{2}$$

Divide 4 by 2.

$$q=2$$

Now solve the equation q=\frac{-7±11}{2} when ± is minus. Subtract 11 from -7.

$$q=-\frac{18}{2}$$

Divide -18 by 2.

$$q=-9$$

The equation is now solved.

$$q=2 q=-9$$

Steps for Completing the Square

Subtract 3q^{2} from both sides.

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

Combine 4q^{2} and -3q^{2} to get q^{2}.

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

Add 4q to both sides.

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

Combine 3q and 4q to get 7q.

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

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}+7q+\left(\frac{7}{2}\right)^{2}=18+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$q^{2}+7q+\frac{49}{4}=18+\frac{49}{4}$$

Add 18 to \frac{49}{4}.

$$q^{2}+7q+\frac{49}{4}=\frac{121}{4}$$

Factor q^{2}+7q+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+\frac{7}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+\frac{7}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$q+\frac{7}{2}=\frac{11}{2} q+\frac{7}{2}=-\frac{11}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$q=2 q=-9$$