Solve 0=-x^2+10x-24

$$0 = - { x }^{ 2 } +10x-24$$

Solve for x

$$x=4$$

$$x=6$$

Steps Using Factoring By Grouping

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

$$-x^{2}+10x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-24\right)=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=6 b=4$$

Rewrite -x^{2}+10x-24 as \left(-x^{2}+6x\right)+\left(4x-24\right).

$$\left(-x^{2}+6x\right)+\left(4x-24\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-6\right)+4\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+4\right)$$

To find equation solutions, solve x-6=0 and -x+4=0.

$$x=6 x=4$$

Steps Using the Quadratic Formula

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

$$-x^{2}+10x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-24\right)}}{2\left(-1\right)}$$

Multiply 4 times -24.

$$x=\frac{-10±\sqrt{100-96}}{2\left(-1\right)}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±2}{-2}$$

Now solve the equation x=\frac{-10±2}{-2} when ± is plus. Add -10 to 2.

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

Divide -8 by -2.

$$x=4$$

Now solve the equation x=\frac{-10±2}{-2} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=4 x=6$$

Steps for Completing the Square

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

$$-x^{2}+10x-24=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$-x^{2}+10x=24$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{24}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{24}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{24}{-1}$$

Divide 24 by -1.

$$x^{2}-10x=-24$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-24+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-24+25$$

Add -24 to 25.

$$x^{2}-10x+25=1$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-5=1 x-5=-1$$

Add 5 to both sides of the equation.

$$x=6 x=4$$