Solve 18=-2s^2+12s

$$18 = - 2 s ^ { 2 } + 12 s$$

Solve for s

$$s=3$$

Steps Using Factoring By Grouping

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

$$-2s^{2}+12s=18$$

Subtract 18 from both sides.

$$-2s^{2}+12s-18=0$$

Divide both sides by 2.

$$-s^{2}+6s-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -s^{2}+as+bs-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-9\right)=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -s^{2}+6s-9 as \left(-s^{2}+3s\right)+\left(3s-9\right).

$$\left(-s^{2}+3s\right)+\left(3s-9\right)$$

Factor out -s in the first and 3 in the second group.

$$-s\left(s-3\right)+3\left(s-3\right)$$

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

$$\left(s-3\right)\left(-s+3\right)$$

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

$$s=3 s=3$$

Steps Using the Quadratic Formula

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

$$-2s^{2}+12s=18$$

Subtract 18 from both sides.

$$-2s^{2}+12s-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Square 12.

$$s=\frac{-12±\sqrt{144-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$s=\frac{-12±\sqrt{144+8\left(-18\right)}}{2\left(-2\right)}$$

Multiply 8 times -18.

$$s=\frac{-12±\sqrt{144-144}}{2\left(-2\right)}$$

Add 144 to -144.

$$s=\frac{-12±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$s=-\frac{12}{2\left(-2\right)}$$

Multiply 2 times -2.

$$s=-\frac{12}{-4}$$

Divide -12 by -4.

$$s=3$$

Steps for Completing the Square

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

$$-2s^{2}+12s=18$$

Divide both sides by -2.

$$\frac{-2s^{2}+12s}{-2}=\frac{18}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$s^{2}+\frac{12}{-2}s=\frac{18}{-2}$$

Divide 12 by -2.

$$s^{2}-6s=\frac{18}{-2}$$

Divide 18 by -2.

$$s^{2}-6s=-9$$

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.

$$s^{2}-6s+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$s^{2}-6s+9=-9+9$$

Add -9 to 9.

$$s^{2}-6s+9=0$$

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

$$\left(s-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$s-3=0 s-3=0$$

Add 3 to both sides of the equation.

$$s=3 s=3$$

The equation is now solved. Solutions are the same.

$$s=3$$