Solve 2a^2-10a+29+9=50

$$2 a ^ { 2 } - 10 a + 29 + 9 = 50$$

Solve for a

$$a=-1$$

$$a=6$$

Steps Using Factoring By Grouping

Add 29 and 9 to get 38.

$$2a^{2}-10a+38=50$$

Subtract 50 from both sides.

$$2a^{2}-10a+38-50=0$$

Subtract 50 from 38 to get -12.

$$2a^{2}-10a-12=0$$

Divide both sides by 2.

$$a^{2}-5a-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite a^{2}-5a-6 as \left(a^{2}-6a\right)+\left(a-6\right).

$$\left(a^{2}-6a\right)+\left(a-6\right)$$

Factor out a in a^{2}-6a.

$$a\left(a-6\right)+a-6$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(a+1\right)$$

To find equation solutions, solve a-6=0 and a+1=0.

$$a=6 a=-1$$

Steps Using the Quadratic Formula

Add 29 and 9 to get 38.

$$2a^{2}-10a+38=50$$

Subtract 50 from both sides.

$$2a^{2}-10a+38-50=0$$

Subtract 50 from 38 to get -12.

$$2a^{2}-10a-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square -10.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-10\right)±\sqrt{100-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$a=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 2}$$

Add 100 to 96.

$$a=\frac{-\left(-10\right)±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$a=\frac{-\left(-10\right)±14}{2\times 2}$$

The opposite of -10 is 10.

$$a=\frac{10±14}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{10±14}{4}$$

Now solve the equation a=\frac{10±14}{4} when ± is plus. Add 10 to 14.

$$a=\frac{24}{4}$$

Divide 24 by 4.

$$a=6$$

Now solve the equation a=\frac{10±14}{4} when ± is minus. Subtract 14 from 10.

$$a=-\frac{4}{4}$$

Divide -4 by 4.

$$a=-1$$

The equation is now solved.

$$a=6 a=-1$$

Steps for Completing the Square

Add 29 and 9 to get 38.

$$2a^{2}-10a+38=50$$

Subtract 38 from both sides.

$$2a^{2}-10a=50-38$$

Subtract 38 from 50 to get 12.

$$2a^{2}-10a=12$$

Divide both sides by 2.

$$\frac{2a^{2}-10a}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}+\left(-\frac{10}{2}\right)a=\frac{12}{2}$$

Divide -10 by 2.

$$a^{2}-5a=\frac{12}{2}$$

Divide 12 by 2.

$$a^{2}-5a=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-5a+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-5a+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$a^{2}-5a+\frac{25}{4}=\frac{49}{4}$$

Factor a^{2}-5a+\frac{25}{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}.

$$\left(a-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$a-\frac{5}{2}=\frac{7}{2} a-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$a=6 a=-1$$