Solve 5x=x^2+25/2

$$5 x = \frac { x ^ { 2 } + 25 } { 2 }$$

Solve for x

$$x=5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

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

Subtract x^{2} from both sides.

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

Subtract 25 from both sides.

$$10x-x^{2}-25=0$$

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

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

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

$$a+b=10 ab=-\left(-25\right)=25$$

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

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

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

$$-x\left(x-5\right)+5\left(x-5\right)$$

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

$$\left(x-5\right)\left(-x+5\right)$$

To find equation solutions, solve x-5=0 and -x+5=0.

$$x=5 x=5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Subtract x^{2} from both sides.

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

Subtract 25 from both sides.

$$10x-x^{2}-25=0$$

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.

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

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}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-25\right)}}{2\left(-1\right)}$$

Multiply 4 times -25.

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

Add 100 to -100.

$$x=\frac{-10±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

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

Multiply 2 times -1.

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

Divide -10 by -2.

$$x=5$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Subtract x^{2} from both sides.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 10 by -1.

$$x^{2}-10x=\frac{25}{-1}$$

Divide 25 by -1.

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

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.

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

Square -5.

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

Add -25 to 25.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$