Solve 5+(c-1)^2+5+(c+4)^2=35

$$5 + ( c - 1 ) ^ { 2 } + 5 + ( c + 4 ) ^ { 2 } = 35$$

Solve for c

$$c=-4$$

$$c=1$$

Steps Using Factoring By Grouping

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

$$5+c^{2}-2c+1+5+\left(c+4\right)^{2}=35$$

Add 5 and 1 to get 6.

$$6+c^{2}-2c+5+\left(c+4\right)^{2}=35$$

Add 6 and 5 to get 11.

$$11+c^{2}-2c+\left(c+4\right)^{2}=35$$

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

$$11+c^{2}-2c+c^{2}+8c+16=35$$

Combine c^{2} and c^{2} to get 2c^{2}.

$$11+2c^{2}-2c+8c+16=35$$

Combine -2c and 8c to get 6c.

$$11+2c^{2}+6c+16=35$$

Add 11 and 16 to get 27.

$$27+2c^{2}+6c=35$$

Subtract 35 from both sides.

$$27+2c^{2}+6c-35=0$$

Subtract 35 from 27 to get -8.

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

Divide both sides by 2.

$$-4+c^{2}+3c=0$$

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

$$c^{2}+3c-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as c^{2}+ac+bc-4. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-4\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite c^{2}+3c-4 as \left(c^{2}-c\right)+\left(4c-4\right).

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

Factor out c in the first and 4 in the second group.

$$c\left(c-1\right)+4\left(c-1\right)$$

Factor out common term c-1 by using distributive property.

$$\left(c-1\right)\left(c+4\right)$$

To find equation solutions, solve c-1=0 and c+4=0.

$$c=1 c=-4$$

Steps Using the Quadratic Formula

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

$$5+c^{2}-2c+1+5+\left(c+4\right)^{2}=35$$

Add 5 and 1 to get 6.

$$6+c^{2}-2c+5+\left(c+4\right)^{2}=35$$

Add 6 and 5 to get 11.

$$11+c^{2}-2c+\left(c+4\right)^{2}=35$$

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

$$11+c^{2}-2c+c^{2}+8c+16=35$$

Combine c^{2} and c^{2} to get 2c^{2}.

$$11+2c^{2}-2c+8c+16=35$$

Combine -2c and 8c to get 6c.

$$11+2c^{2}+6c+16=35$$

Add 11 and 16 to get 27.

$$27+2c^{2}+6c=35$$

Subtract 35 from both sides.

$$27+2c^{2}+6c-35=0$$

Subtract 35 from 27 to get -8.

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

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.

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

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

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

Square 6.

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

Multiply -4 times 2.

$$c=\frac{-6±\sqrt{36-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$c=\frac{-6±\sqrt{36+64}}{2\times 2}$$

Add 36 to 64.

$$c=\frac{-6±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$c=\frac{-6±10}{2\times 2}$$

Multiply 2 times 2.

$$c=\frac{-6±10}{4}$$

Now solve the equation c=\frac{-6±10}{4} when ± is plus. Add -6 to 10.

$$c=\frac{4}{4}$$

Divide 4 by 4.

$$c=1$$

Now solve the equation c=\frac{-6±10}{4} when ± is minus. Subtract 10 from -6.

$$c=-\frac{16}{4}$$

Divide -16 by 4.

$$c=-4$$

The equation is now solved.

$$c=1 c=-4$$

Steps for Completing the Square

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

$$5+c^{2}-2c+1+5+\left(c+4\right)^{2}=35$$

Add 5 and 1 to get 6.

$$6+c^{2}-2c+5+\left(c+4\right)^{2}=35$$

Add 6 and 5 to get 11.

$$11+c^{2}-2c+\left(c+4\right)^{2}=35$$

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

$$11+c^{2}-2c+c^{2}+8c+16=35$$

Combine c^{2} and c^{2} to get 2c^{2}.

$$11+2c^{2}-2c+8c+16=35$$

Combine -2c and 8c to get 6c.

$$11+2c^{2}+6c+16=35$$

Add 11 and 16 to get 27.

$$27+2c^{2}+6c=35$$

Subtract 27 from both sides.

$$2c^{2}+6c=35-27$$

Subtract 27 from 35 to get 8.

$$2c^{2}+6c=8$$

Divide both sides by 2.

$$\frac{2c^{2}+6c}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$c^{2}+\frac{6}{2}c=\frac{8}{2}$$

Divide 6 by 2.

$$c^{2}+3c=\frac{8}{2}$$

Divide 8 by 2.

$$c^{2}+3c=4$$

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

$$c^{2}+3c+\left(\frac{3}{2}\right)^{2}=4+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$c^{2}+3c+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$c^{2}+3c+\frac{9}{4}=\frac{25}{4}$$

Factor c^{2}+3c+\frac{9}{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}.

$$\left(c+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$c+\frac{3}{2}=\frac{5}{2} c+\frac{3}{2}=-\frac{5}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$c=1 c=-4$$