Solve 4{left({a}_{2}right)}^2=left(5-{a}_{2}right)left(21-{a}_{2}right)

$$4 { \left( { a }_{ 2 } \right) }^{ 2 } = \left( 5- { a }_{ 2 } \right) \left( 21- { a }_{ 2 } \right)$$

Solve for a_2

$$a_{2} = -\frac{35}{3} = -11\frac{2}{3} \approx -11.666666667$$

$$a_{2}=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 5-a_{2} by 21-a_{2} and combine like terms.

$$4a_{2}^{2}=105-26a_{2}+a_{2}^{2}$$

Subtract 105 from both sides.

$$4a_{2}^{2}-105=-26a_{2}+a_{2}^{2}$$

Add 26a_{2} to both sides.

$$4a_{2}^{2}-105+26a_{2}=a_{2}^{2}$$

Subtract a_{2}^{2} from both sides.

$$4a_{2}^{2}-105+26a_{2}-a_{2}^{2}=0$$

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

$$3a_{2}^{2}-105+26a_{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3a_{2}^{2}+26a_{2}-105=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3a_{2}^{2}+aa_{2}+ba_{2}-105. To find a and b, set up a system to be solved.

$$a+b=26 ab=3\left(-105\right)=-315$$

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 -315.

$$-1,315 -3,105 -5,63 -7,45 -9,35 -15,21$$

Calculate the sum for each pair.

$$-1+315=314 -3+105=102 -5+63=58 -7+45=38 -9+35=26 -15+21=6$$

The solution is the pair that gives sum 26.

$$a=-9 b=35$$

Rewrite 3a_{2}^{2}+26a_{2}-105 as \left(3a_{2}^{2}-9a_{2}\right)+\left(35a_{2}-105\right).

$$\left(3a_{2}^{2}-9a_{2}\right)+\left(35a_{2}-105\right)$$

Factor out 3a_{2} in the first and 35 in the second group.

$$3a_{2}\left(a_{2}-3\right)+35\left(a_{2}-3\right)$$

Factor out common term a_{2}-3 by using distributive property.

$$\left(a_{2}-3\right)\left(3a_{2}+35\right)$$

To find equation solutions, solve a_{2}-3=0 and 3a_{2}+35=0.

$$a_{2}=3 a_{2}=-\frac{35}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5-a_{2} by 21-a_{2} and combine like terms.

$$4a_{2}^{2}=105-26a_{2}+a_{2}^{2}$$

Subtract 105 from both sides.

$$4a_{2}^{2}-105=-26a_{2}+a_{2}^{2}$$

Add 26a_{2} to both sides.

$$4a_{2}^{2}-105+26a_{2}=a_{2}^{2}$$

Subtract a_{2}^{2} from both sides.

$$4a_{2}^{2}-105+26a_{2}-a_{2}^{2}=0$$

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

$$3a_{2}^{2}-105+26a_{2}=0$$

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.

$$3a_{2}^{2}+26a_{2}-105=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 26 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a_{2}=\frac{-26±\sqrt{26^{2}-4\times 3\left(-105\right)}}{2\times 3}$$

Square 26.

$$a_{2}=\frac{-26±\sqrt{676-4\times 3\left(-105\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a_{2}=\frac{-26±\sqrt{676-12\left(-105\right)}}{2\times 3}$$

Multiply -12 times -105.

$$a_{2}=\frac{-26±\sqrt{676+1260}}{2\times 3}$$

Add 676 to 1260.

$$a_{2}=\frac{-26±\sqrt{1936}}{2\times 3}$$

Take the square root of 1936.

$$a_{2}=\frac{-26±44}{2\times 3}$$

Multiply 2 times 3.

$$a_{2}=\frac{-26±44}{6}$$

Now solve the equation a_{2}=\frac{-26±44}{6} when ± is plus. Add -26 to 44.

$$a_{2}=\frac{18}{6}$$

Divide 18 by 6.

$$a_{2}=3$$

Now solve the equation a_{2}=\frac{-26±44}{6} when ± is minus. Subtract 44 from -26.

$$a_{2}=-\frac{70}{6}$$

Reduce the fraction \frac{-70}{6} to lowest terms by extracting and canceling out 2.

$$a_{2}=-\frac{35}{3}$$

The equation is now solved.

$$a_{2}=3 a_{2}=-\frac{35}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 5-a_{2} by 21-a_{2} and combine like terms.

$$4a_{2}^{2}=105-26a_{2}+a_{2}^{2}$$

Add 26a_{2} to both sides.

$$4a_{2}^{2}+26a_{2}=105+a_{2}^{2}$$

Subtract a_{2}^{2} from both sides.

$$4a_{2}^{2}+26a_{2}-a_{2}^{2}=105$$

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

$$3a_{2}^{2}+26a_{2}=105$$

Divide both sides by 3.

$$\frac{3a_{2}^{2}+26a_{2}}{3}=\frac{105}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a_{2}^{2}+\frac{26}{3}a_{2}=\frac{105}{3}$$

Divide 105 by 3.

$$a_{2}^{2}+\frac{26}{3}a_{2}=35$$

Divide \frac{26}{3}, the coefficient of the x term, by 2 to get \frac{13}{3}. Then add the square of \frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a_{2}^{2}+\frac{26}{3}a_{2}+\left(\frac{13}{3}\right)^{2}=35+\left(\frac{13}{3}\right)^{2}$$

Square \frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$a_{2}^{2}+\frac{26}{3}a_{2}+\frac{169}{9}=35+\frac{169}{9}$$

Add 35 to \frac{169}{9}.

$$a_{2}^{2}+\frac{26}{3}a_{2}+\frac{169}{9}=\frac{484}{9}$$

Factor a_{2}^{2}+\frac{26}{3}a_{2}+\frac{169}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a_{2}+\frac{13}{3}\right)^{2}=\frac{484}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a_{2}+\frac{13}{3}\right)^{2}}=\sqrt{\frac{484}{9}}$$

Simplify.

$$a_{2}+\frac{13}{3}=\frac{22}{3} a_{2}+\frac{13}{3}=-\frac{22}{3}$$

Subtract \frac{13}{3} from both sides of the equation.

$$a_{2}=3 a_{2}=-\frac{35}{3}$$