Solve -3+2a=left(4-aright)^2-4left(4-aright)+5

$$-3+2a = { \left(4-a \right) }^{ 2 } -4 \left( 4-a \right) +5$$

Solve for a

$$a=2$$

$$a=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$-3+2a=16-8a+a^{2}-4\left(4-a\right)+5$$

Use the distributive property to multiply -4 by 4-a.

$$-3+2a=16-8a+a^{2}-16+4a+5$$

Subtract 16 from 16 to get 0.

$$-3+2a=-8a+a^{2}+4a+5$$

Combine -8a and 4a to get -4a.

$$-3+2a=-4a+a^{2}+5$$

Add 4a to both sides.

$$-3+2a+4a=a^{2}+5$$

Combine 2a and 4a to get 6a.

$$-3+6a=a^{2}+5$$

Subtract a^{2} from both sides.

$$-3+6a-a^{2}=5$$

Subtract 5 from both sides.

$$-3+6a-a^{2}-5=0$$

Subtract 5 from -3 to get -8.

$$-8+6a-a^{2}=0$$

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

$$-a^{2}+6a-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -a^{2}+6a-8 as \left(-a^{2}+4a\right)+\left(2a-8\right).

$$\left(-a^{2}+4a\right)+\left(2a-8\right)$$

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

$$-a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(-a+2\right)$$

To find equation solutions, solve a-4=0 and -a+2=0.

$$a=4 a=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$-3+2a=16-8a+a^{2}-4\left(4-a\right)+5$$

Use the distributive property to multiply -4 by 4-a.

$$-3+2a=16-8a+a^{2}-16+4a+5$$

Subtract 16 from 16 to get 0.

$$-3+2a=-8a+a^{2}+4a+5$$

Combine -8a and 4a to get -4a.

$$-3+2a=-4a+a^{2}+5$$

Add 4a to both sides.

$$-3+2a+4a=a^{2}+5$$

Combine 2a and 4a to get 6a.

$$-3+6a=a^{2}+5$$

Subtract a^{2} from both sides.

$$-3+6a-a^{2}=5$$

Subtract 5 from both sides.

$$-3+6a-a^{2}-5=0$$

Subtract 5 from -3 to get -8.

$$-8+6a-a^{2}=0$$

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.

$$-a^{2}+6a-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square 6.

$$a=\frac{-6±\sqrt{36-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-6±\sqrt{36+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply 4 times -8.

$$a=\frac{-6±\sqrt{36-32}}{2\left(-1\right)}$$

Add 36 to -32.

$$a=\frac{-6±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$a=\frac{-6±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-6±2}{-2}$$

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

$$a=-\frac{4}{-2}$$

Divide -4 by -2.

$$a=2$$

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

$$a=-\frac{8}{-2}$$

Divide -8 by -2.

$$a=4$$

The equation is now solved.

$$a=2 a=4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$-3+2a=16-8a+a^{2}-4\left(4-a\right)+5$$

Use the distributive property to multiply -4 by 4-a.

$$-3+2a=16-8a+a^{2}-16+4a+5$$

Subtract 16 from 16 to get 0.

$$-3+2a=-8a+a^{2}+4a+5$$

Combine -8a and 4a to get -4a.

$$-3+2a=-4a+a^{2}+5$$

Add 4a to both sides.

$$-3+2a+4a=a^{2}+5$$

Combine 2a and 4a to get 6a.

$$-3+6a=a^{2}+5$$

Subtract a^{2} from both sides.

$$-3+6a-a^{2}=5$$

Add 3 to both sides.

$$6a-a^{2}=5+3$$

Add 5 and 3 to get 8.

$$6a-a^{2}=8$$

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.

$$-a^{2}+6a=8$$

Divide both sides by -1.

$$\frac{-a^{2}+6a}{-1}=\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\frac{6}{-1}a=\frac{8}{-1}$$

Divide 6 by -1.

$$a^{2}-6a=\frac{8}{-1}$$

Divide 8 by -1.

$$a^{2}-6a=-8$$

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

$$a^{2}-6a+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$a^{2}-6a+9=-8+9$$

Add -8 to 9.

$$a^{2}-6a+9=1$$

Factor a^{2}-6a+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}.

$$\left(a-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$a-3=1 a-3=-1$$

Add 3 to both sides of the equation.

$$a=4 a=2$$