Solve 16a^2+21a+9=6a^2

$$16 a ^ { 2 } + 21 a + 9 = 6 a ^ { 2 }$$

Solve for a

$$a=-\frac{3}{5}=-0.6$$

$$a = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

Subtract 6a^{2} from both sides.

$$16a^{2}+21a+9-6a^{2}=0$$

Combine 16a^{2} and -6a^{2} to get 10a^{2}.

$$10a^{2}+21a+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10a^{2}+aa+ba+9. To find a and b, set up a system to be solved.

$$a+b=21 ab=10\times 9=90$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 21.

$$a=6 b=15$$

Rewrite 10a^{2}+21a+9 as \left(10a^{2}+6a\right)+\left(15a+9\right).

$$\left(10a^{2}+6a\right)+\left(15a+9\right)$$

Factor out 2a in the first and 3 in the second group.

$$2a\left(5a+3\right)+3\left(5a+3\right)$$

Factor out common term 5a+3 by using distributive property.

$$\left(5a+3\right)\left(2a+3\right)$$

To find equation solutions, solve 5a+3=0 and 2a+3=0.

$$a=-\frac{3}{5} a=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Subtract 6a^{2} from both sides.

$$16a^{2}+21a+9-6a^{2}=0$$

Combine 16a^{2} and -6a^{2} to get 10a^{2}.

$$10a^{2}+21a+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 21 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-21±\sqrt{21^{2}-4\times 10\times 9}}{2\times 10}$$

Square 21.

$$a=\frac{-21±\sqrt{441-4\times 10\times 9}}{2\times 10}$$

Multiply -4 times 10.

$$a=\frac{-21±\sqrt{441-40\times 9}}{2\times 10}$$

Multiply -40 times 9.

$$a=\frac{-21±\sqrt{441-360}}{2\times 10}$$

Add 441 to -360.

$$a=\frac{-21±\sqrt{81}}{2\times 10}$$

Take the square root of 81.

$$a=\frac{-21±9}{2\times 10}$$

Multiply 2 times 10.

$$a=\frac{-21±9}{20}$$

Now solve the equation a=\frac{-21±9}{20} when ± is plus. Add -21 to 9.

$$a=-\frac{12}{20}$$

Reduce the fraction \frac{-12}{20} to lowest terms by extracting and canceling out 4.

$$a=-\frac{3}{5}$$

Now solve the equation a=\frac{-21±9}{20} when ± is minus. Subtract 9 from -21.

$$a=-\frac{30}{20}$$

Reduce the fraction \frac{-30}{20} to lowest terms by extracting and canceling out 10.

$$a=-\frac{3}{2}$$

The equation is now solved.

$$a=-\frac{3}{5} a=-\frac{3}{2}$$

Steps for Completing the Square

Subtract 6a^{2} from both sides.

$$16a^{2}+21a+9-6a^{2}=0$$

Combine 16a^{2} and -6a^{2} to get 10a^{2}.

$$10a^{2}+21a+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$10a^{2}+21a=-9$$

Divide both sides by 10.

$$\frac{10a^{2}+21a}{10}=-\frac{9}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$a^{2}+\frac{21}{10}a=-\frac{9}{10}$$

Divide \frac{21}{10}, the coefficient of the x term, by 2 to get \frac{21}{20}. Then add the square of \frac{21}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{21}{10}a+\left(\frac{21}{20}\right)^{2}=-\frac{9}{10}+\left(\frac{21}{20}\right)^{2}$$

Square \frac{21}{20} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{21}{10}a+\frac{441}{400}=-\frac{9}{10}+\frac{441}{400}$$

Add -\frac{9}{10} to \frac{441}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{21}{10}a+\frac{441}{400}=\frac{81}{400}$$

Factor a^{2}+\frac{21}{10}a+\frac{441}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{21}{20}\right)^{2}=\frac{81}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{21}{20}\right)^{2}}=\sqrt{\frac{81}{400}}$$

Simplify.

$$a+\frac{21}{20}=\frac{9}{20} a+\frac{21}{20}=-\frac{9}{20}$$

Subtract \frac{21}{20} from both sides of the equation.

$$a=-\frac{3}{5} a=-\frac{3}{2}$$