Solve r^2+25=10r

$$r ^ { 2 } + 25 = 10 r$$

Solve for r

$$r=5$$

Steps Using Factoring

Subtract 10r from both sides.

$$r^{2}+25-10r=0$$

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

$$r^{2}-10r+25=0$$

To solve the equation, factor r^{2}-10r+25 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.

$$\left(r-5\right)\left(r-5\right)$$

Rewrite as a binomial square.

$$\left(r-5\right)^{2}$$

To find equation solution, solve r-5=0.

$$r=5$$

Steps Using Factoring By Grouping

Subtract 10r from both sides.

$$r^{2}+25-10r=0$$

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

$$r^{2}-10r+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br+25. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 25=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite r^{2}-10r+25 as \left(r^{2}-5r\right)+\left(-5r+25\right).

$$\left(r^{2}-5r\right)+\left(-5r+25\right)$$

Factor out r in the first and -5 in the second group.

$$r\left(r-5\right)-5\left(r-5\right)$$

Factor out common term r-5 by using distributive property.

$$\left(r-5\right)\left(r-5\right)$$

Rewrite as a binomial square.

$$\left(r-5\right)^{2}$$

To find equation solution, solve r-5=0.

$$r=5$$

Steps Using the Quadratic Formula

Subtract 10r from both sides.

$$r^{2}+25-10r=0$$

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.

$$r^{2}-10r+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}$$

Square -10.

$$r=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}$$

Multiply -4 times 25.

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

Add 100 to -100.

$$r=\frac{-\left(-10\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$r=-\frac{-10}{2}$$

The opposite of -10 is 10.

$$r=\frac{10}{2}$$

Divide 10 by 2.

$$r=5$$

Steps for Completing the Square

Subtract 10r from both sides.

$$r^{2}+25-10r=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$r^{2}-10r=-25$$

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.

$$r^{2}-10r+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Square -5.

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

Add -25 to 25.

$$r^{2}-10r+25=0$$

Factor r^{2}-10r+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}.

$$\left(r-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$r-5=0 r-5=0$$

Add 5 to both sides of the equation.

$$r=5 r=5$$

The equation is now solved. Solutions are the same.

$$r=5$$