Solve 5c-c^2=25-10c+2c^2-7

$$5 c - c ^ { 2 } = 25 - 10 c + 2 c ^ { 2 } - 7$$

Solve for c

$$c=2$$

$$c=3$$

Steps Using Factoring By Grouping

Subtract 7 from 25 to get 18.

$$5c-c^{2}=18-10c+2c^{2}$$

Subtract 18 from both sides.

$$5c-c^{2}-18=-10c+2c^{2}$$

Add 10c to both sides.

$$5c-c^{2}-18+10c=2c^{2}$$

Combine 5c and 10c to get 15c.

$$15c-c^{2}-18=2c^{2}$$

Subtract 2c^{2} from both sides.

$$15c-c^{2}-18-2c^{2}=0$$

Combine -c^{2} and -2c^{2} to get -3c^{2}.

$$15c-3c^{2}-18=0$$

Divide both sides by 3.

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

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

$$-c^{2}+5c-6=0$$

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

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

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -c^{2}+5c-6 as \left(-c^{2}+3c\right)+\left(2c-6\right).

$$\left(-c^{2}+3c\right)+\left(2c-6\right)$$

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

$$-c\left(c-3\right)+2\left(c-3\right)$$

Factor out common term c-3 by using distributive property.

$$\left(c-3\right)\left(-c+2\right)$$

To find equation solutions, solve c-3=0 and -c+2=0.

$$c=3 c=2$$

Steps Using the Quadratic Formula

Subtract 7 from 25 to get 18.

$$5c-c^{2}=18-10c+2c^{2}$$

Subtract 18 from both sides.

$$5c-c^{2}-18=-10c+2c^{2}$$

Add 10c to both sides.

$$5c-c^{2}-18+10c=2c^{2}$$

Combine 5c and 10c to get 15c.

$$15c-c^{2}-18=2c^{2}$$

Subtract 2c^{2} from both sides.

$$15c-c^{2}-18-2c^{2}=0$$

Combine -c^{2} and -2c^{2} to get -3c^{2}.

$$15c-3c^{2}-18=0$$

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.

$$-3c^{2}+15c-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 15 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-15±\sqrt{15^{2}-4\left(-3\right)\left(-18\right)}}{2\left(-3\right)}$$

Square 15.

$$c=\frac{-15±\sqrt{225-4\left(-3\right)\left(-18\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$c=\frac{-15±\sqrt{225+12\left(-18\right)}}{2\left(-3\right)}$$

Multiply 12 times -18.

$$c=\frac{-15±\sqrt{225-216}}{2\left(-3\right)}$$

Add 225 to -216.

$$c=\frac{-15±\sqrt{9}}{2\left(-3\right)}$$

Take the square root of 9.

$$c=\frac{-15±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$c=\frac{-15±3}{-6}$$

Now solve the equation c=\frac{-15±3}{-6} when ± is plus. Add -15 to 3.

$$c=-\frac{12}{-6}$$

Divide -12 by -6.

$$c=2$$

Now solve the equation c=\frac{-15±3}{-6} when ± is minus. Subtract 3 from -15.

$$c=-\frac{18}{-6}$$

Divide -18 by -6.

$$c=3$$

The equation is now solved.

$$c=2 c=3$$

Steps for Completing the Square

Subtract 7 from 25 to get 18.

$$5c-c^{2}=18-10c+2c^{2}$$

Add 10c to both sides.

$$5c-c^{2}+10c=18+2c^{2}$$

Combine 5c and 10c to get 15c.

$$15c-c^{2}=18+2c^{2}$$

Subtract 2c^{2} from both sides.

$$15c-c^{2}-2c^{2}=18$$

Combine -c^{2} and -2c^{2} to get -3c^{2}.

$$15c-3c^{2}=18$$

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.

$$-3c^{2}+15c=18$$

Divide both sides by -3.

$$\frac{-3c^{2}+15c}{-3}=\frac{18}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$c^{2}+\frac{15}{-3}c=\frac{18}{-3}$$

Divide 15 by -3.

$$c^{2}-5c=\frac{18}{-3}$$

Divide 18 by -3.

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

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.

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

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

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

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

$$c^{2}-5c+\frac{25}{4}=\frac{1}{4}$$

Factor c^{2}-5c+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$c-\frac{5}{2}=\frac{1}{2} c-\frac{5}{2}=-\frac{1}{2}$$

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

$$c=3 c=2$$