Solve 26x=x^2+16x

$$26 x = x ^ { 2 } + 16 x$$

Solve for x

$$x=10$$

$$x=0$$

Steps Using Factoring

Subtract x^{2} from both sides.

$$26x-x^{2}=16x$$

Subtract 16x from both sides.

$$26x-x^{2}-16x=0$$

Combine 26x and -16x to get 10x.

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

Factor out x.

$$x\left(10-x\right)=0$$

To find equation solutions, solve x=0 and 10-x=0.

$$x=0 x=10$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$26x-x^{2}=16x$$

Subtract 16x from both sides.

$$26x-x^{2}-16x=0$$

Combine 26x and -16x to get 10x.

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

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.

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

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

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

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±10}{-2}$$

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

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

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

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

The equation is now solved.

$$x=0 x=10$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$26x-x^{2}=16x$$

Subtract 16x from both sides.

$$26x-x^{2}-16x=0$$

Combine 26x and -16x to get 10x.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{0}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{0}{-1}$$

Divide 0 by -1.

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

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.

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

Square -5.

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

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}.

$$\left(x-5\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{25}$$

Simplify.

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

Add 5 to both sides of the equation.

$$x=10 x=0$$