Solve 3a^2-29a+61=5

$$3 a ^ { 2 } - 29 a + 61 = 5$$

Solve for a

$$a=7$$

$$a = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Subtract 5 from both sides.

$$3a^{2}-29a+61-5=0$$

Subtract 5 from 61 to get 56.

$$3a^{2}-29a+56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3a^{2}+aa+ba+56. To find a and b, set up a system to be solved.

$$a+b=-29 ab=3\times 56=168$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 168.

$$-1,-168 -2,-84 -3,-56 -4,-42 -6,-28 -7,-24 -8,-21 -12,-14$$

Calculate the sum for each pair.

$$-1-168=-169 -2-84=-86 -3-56=-59 -4-42=-46 -6-28=-34 -7-24=-31 -8-21=-29 -12-14=-26$$

The solution is the pair that gives sum -29.

$$a=-21 b=-8$$

Rewrite 3a^{2}-29a+56 as \left(3a^{2}-21a\right)+\left(-8a+56\right).

$$\left(3a^{2}-21a\right)+\left(-8a+56\right)$$

Factor out 3a in the first and -8 in the second group.

$$3a\left(a-7\right)-8\left(a-7\right)$$

Factor out common term a-7 by using distributive property.

$$\left(a-7\right)\left(3a-8\right)$$

To find equation solutions, solve a-7=0 and 3a-8=0.

$$a=7 a=\frac{8}{3}$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3a^{2}-29a+61=5$$

Subtract 5 from both sides of the equation.

$$3a^{2}-29a+61-5=5-5$$

Subtracting 5 from itself leaves 0.

$$3a^{2}-29a+61-5=0$$

Subtract 5 from 61.

$$3a^{2}-29a+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -29 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 3\times 56}}{2\times 3}$$

Square -29.

$$a=\frac{-\left(-29\right)±\sqrt{841-4\times 3\times 56}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-29\right)±\sqrt{841-12\times 56}}{2\times 3}$$

Multiply -12 times 56.

$$a=\frac{-\left(-29\right)±\sqrt{841-672}}{2\times 3}$$

Add 841 to -672.

$$a=\frac{-\left(-29\right)±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$a=\frac{-\left(-29\right)±13}{2\times 3}$$

The opposite of -29 is 29.

$$a=\frac{29±13}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{29±13}{6}$$

Now solve the equation a=\frac{29±13}{6} when ± is plus. Add 29 to 13.

$$a=\frac{42}{6}$$

Divide 42 by 6.

$$a=7$$

Now solve the equation a=\frac{29±13}{6} when ± is minus. Subtract 13 from 29.

$$a=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$a=\frac{8}{3}$$

The equation is now solved.

$$a=7 a=\frac{8}{3}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3a^{2}-29a+61=5$$

Subtract 61 from both sides of the equation.

$$3a^{2}-29a+61-61=5-61$$

Subtracting 61 from itself leaves 0.

$$3a^{2}-29a=5-61$$

Subtract 61 from 5.

$$3a^{2}-29a=-56$$

Divide both sides by 3.

$$\frac{3a^{2}-29a}{3}=-\frac{56}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a^{2}-\frac{29}{3}a=-\frac{56}{3}$$

Divide -\frac{29}{3}, the coefficient of the x term, by 2 to get -\frac{29}{6}. Then add the square of -\frac{29}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{29}{3}a+\left(-\frac{29}{6}\right)^{2}=-\frac{56}{3}+\left(-\frac{29}{6}\right)^{2}$$

Square -\frac{29}{6} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{29}{3}a+\frac{841}{36}=-\frac{56}{3}+\frac{841}{36}$$

Add -\frac{56}{3} to \frac{841}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{29}{3}a+\frac{841}{36}=\frac{169}{36}$$

Factor a^{2}-\frac{29}{3}a+\frac{841}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{29}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{29}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$a-\frac{29}{6}=\frac{13}{6} a-\frac{29}{6}=-\frac{13}{6}$$

Add \frac{29}{6} to both sides of the equation.

$$a=7 a=\frac{8}{3}$$