Solve 133=a+(a+1)^2+(a+2-7)

$$133 = a + ( a + 1 ) ^ { 2 } + ( a + 2 - 7 )$$

Solve for a (complex solution)

$$a=\sqrt{141}-2\approx 9.874342087$$

$$a=-\left(\sqrt{141}+2\right)\approx -13.874342087$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+1\right)^{2}.

$$133=a+a^{2}+2a+1+a+2-7$$

Combine a and 2a to get 3a.

$$133=3a+a^{2}+1+a+2-7$$

Combine 3a and a to get 4a.

$$133=4a+a^{2}+1+2-7$$

Add 1 and 2 to get 3.

$$133=4a+a^{2}+3-7$$

Subtract 7 from 3 to get -4.

$$133=4a+a^{2}-4$$

Swap sides so that all variable terms are on the left hand side.

$$4a+a^{2}-4=133$$

Subtract 133 from both sides.

$$4a+a^{2}-4-133=0$$

Subtract 133 from -4 to get -137.

$$4a+a^{2}-137=0$$

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.

$$a^{2}+4a-137=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -137 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-4±\sqrt{4^{2}-4\left(-137\right)}}{2}$$

Square 4.

$$a=\frac{-4±\sqrt{16-4\left(-137\right)}}{2}$$

Multiply -4 times -137.

$$a=\frac{-4±\sqrt{16+548}}{2}$$

Add 16 to 548.

$$a=\frac{-4±\sqrt{564}}{2}$$

Take the square root of 564.

$$a=\frac{-4±2\sqrt{141}}{2}$$

Now solve the equation a=\frac{-4±2\sqrt{141}}{2} when ± is plus. Add -4 to 2\sqrt{141}.

$$a=\frac{2\sqrt{141}-4}{2}$$

Divide -4+2\sqrt{141} by 2.

$$a=\sqrt{141}-2$$

Now solve the equation a=\frac{-4±2\sqrt{141}}{2} when ± is minus. Subtract 2\sqrt{141} from -4.

$$a=\frac{-2\sqrt{141}-4}{2}$$

Divide -4-2\sqrt{141} by 2.

$$a=-\sqrt{141}-2$$

The equation is now solved.

$$a=\sqrt{141}-2 a=-\sqrt{141}-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+1\right)^{2}.

$$133=a+a^{2}+2a+1+a+2-7$$

Combine a and 2a to get 3a.

$$133=3a+a^{2}+1+a+2-7$$

Combine 3a and a to get 4a.

$$133=4a+a^{2}+1+2-7$$

Add 1 and 2 to get 3.

$$133=4a+a^{2}+3-7$$

Subtract 7 from 3 to get -4.

$$133=4a+a^{2}-4$$

Swap sides so that all variable terms are on the left hand side.

$$4a+a^{2}-4=133$$

Add 4 to both sides.

$$4a+a^{2}=133+4$$

Add 133 and 4 to get 137.

$$4a+a^{2}=137$$

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.

$$a^{2}+4a=137$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+4a+2^{2}=137+2^{2}$$

Square 2.

$$a^{2}+4a+4=137+4$$

Add 137 to 4.

$$a^{2}+4a+4=141$$

Factor a^{2}+4a+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}.

$$\left(a+2\right)^{2}=141$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+2\right)^{2}}=\sqrt{141}$$

Simplify.

$$a+2=\sqrt{141} a+2=-\sqrt{141}$$

Subtract 2 from both sides of the equation.

$$a=\sqrt{141}-2 a=-\sqrt{141}-2$$

Solve for a

$$a=\sqrt{141}-2\approx 9.874342087$$

$$a=-\sqrt{141}-2\approx -13.874342087$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+1\right)^{2}.

$$133=a+a^{2}+2a+1+a+2-7$$

Combine a and 2a to get 3a.

$$133=3a+a^{2}+1+a+2-7$$

Combine 3a and a to get 4a.

$$133=4a+a^{2}+1+2-7$$

Add 1 and 2 to get 3.

$$133=4a+a^{2}+3-7$$

Subtract 7 from 3 to get -4.

$$133=4a+a^{2}-4$$

Swap sides so that all variable terms are on the left hand side.

$$4a+a^{2}-4=133$$

Subtract 133 from both sides.

$$4a+a^{2}-4-133=0$$

Subtract 133 from -4 to get -137.

$$4a+a^{2}-137=0$$

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.

$$a^{2}+4a-137=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -137 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-4±\sqrt{4^{2}-4\left(-137\right)}}{2}$$

Square 4.

$$a=\frac{-4±\sqrt{16-4\left(-137\right)}}{2}$$

Multiply -4 times -137.

$$a=\frac{-4±\sqrt{16+548}}{2}$$

Add 16 to 548.

$$a=\frac{-4±\sqrt{564}}{2}$$

Take the square root of 564.

$$a=\frac{-4±2\sqrt{141}}{2}$$

Now solve the equation a=\frac{-4±2\sqrt{141}}{2} when ± is plus. Add -4 to 2\sqrt{141}.

$$a=\frac{2\sqrt{141}-4}{2}$$

Divide -4+2\sqrt{141} by 2.

$$a=\sqrt{141}-2$$

Now solve the equation a=\frac{-4±2\sqrt{141}}{2} when ± is minus. Subtract 2\sqrt{141} from -4.

$$a=\frac{-2\sqrt{141}-4}{2}$$

Divide -4-2\sqrt{141} by 2.

$$a=-\sqrt{141}-2$$

The equation is now solved.

$$a=\sqrt{141}-2 a=-\sqrt{141}-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+1\right)^{2}.

$$133=a+a^{2}+2a+1+a+2-7$$

Combine a and 2a to get 3a.

$$133=3a+a^{2}+1+a+2-7$$

Combine 3a and a to get 4a.

$$133=4a+a^{2}+1+2-7$$

Add 1 and 2 to get 3.

$$133=4a+a^{2}+3-7$$

Subtract 7 from 3 to get -4.

$$133=4a+a^{2}-4$$

Swap sides so that all variable terms are on the left hand side.

$$4a+a^{2}-4=133$$

Add 4 to both sides.

$$4a+a^{2}=133+4$$

Add 133 and 4 to get 137.

$$4a+a^{2}=137$$

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.

$$a^{2}+4a=137$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+4a+2^{2}=137+2^{2}$$

Square 2.

$$a^{2}+4a+4=137+4$$

Add 137 to 4.

$$a^{2}+4a+4=141$$

Factor a^{2}+4a+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}.

$$\left(a+2\right)^{2}=141$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+2\right)^{2}}=\sqrt{141}$$

Simplify.

$$a+2=\sqrt{141} a+2=-\sqrt{141}$$

Subtract 2 from both sides of the equation.

$$a=\sqrt{141}-2 a=-\sqrt{141}-2$$