Solve 40x=(30-x)^2

$$40 x = ( 30 - x ) ^ { 2 }$$

Solve for x

$$x=10$$

$$x=90$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(30-x\right)^{2}.

$$40x=900-60x+x^{2}$$

Subtract 900 from both sides.

$$40x-900=-60x+x^{2}$$

Add 60x to both sides.

$$40x-900+60x=x^{2}$$

Combine 40x and 60x to get 100x.

$$100x-900=x^{2}$$

Subtract x^{2} from both sides.

$$100x-900-x^{2}=0$$

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

$$-x^{2}+100x-900=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-900. To find a and b, set up a system to be solved.

$$a+b=100 ab=-\left(-900\right)=900$$

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 900.

$$1,900 2,450 3,300 4,225 5,180 6,150 9,100 10,90 12,75 15,60 18,50 20,45 25,36 30,30$$

Calculate the sum for each pair.

$$1+900=901 2+450=452 3+300=303 4+225=229 5+180=185 6+150=156 9+100=109 10+90=100 12+75=87 15+60=75 18+50=68 20+45=65 25+36=61 30+30=60$$

The solution is the pair that gives sum 100.

$$a=90 b=10$$

Rewrite -x^{2}+100x-900 as \left(-x^{2}+90x\right)+\left(10x-900\right).

$$\left(-x^{2}+90x\right)+\left(10x-900\right)$$

Factor out -x in the first and 10 in the second group.

$$-x\left(x-90\right)+10\left(x-90\right)$$

Factor out common term x-90 by using distributive property.

$$\left(x-90\right)\left(-x+10\right)$$

To find equation solutions, solve x-90=0 and -x+10=0.

$$x=90 x=10$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(30-x\right)^{2}.

$$40x=900-60x+x^{2}$$

Subtract 900 from both sides.

$$40x-900=-60x+x^{2}$$

Add 60x to both sides.

$$40x-900+60x=x^{2}$$

Combine 40x and 60x to get 100x.

$$100x-900=x^{2}$$

Subtract x^{2} from both sides.

$$100x-900-x^{2}=0$$

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.

$$-x^{2}+100x-900=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 100 for b, and -900 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-1\right)\left(-900\right)}}{2\left(-1\right)}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-1\right)\left(-900\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-100±\sqrt{10000+4\left(-900\right)}}{2\left(-1\right)}$$

Multiply 4 times -900.

$$x=\frac{-100±\sqrt{10000-3600}}{2\left(-1\right)}$$

Add 10000 to -3600.

$$x=\frac{-100±\sqrt{6400}}{2\left(-1\right)}$$

Take the square root of 6400.

$$x=\frac{-100±80}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-100±80}{-2}$$

Now solve the equation x=\frac{-100±80}{-2} when ± is plus. Add -100 to 80.

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

Divide -20 by -2.

$$x=10$$

Now solve the equation x=\frac{-100±80}{-2} when ± is minus. Subtract 80 from -100.

$$x=-\frac{180}{-2}$$

Divide -180 by -2.

$$x=90$$

The equation is now solved.

$$x=10 x=90$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(30-x\right)^{2}.

$$40x=900-60x+x^{2}$$

Add 60x to both sides.

$$40x+60x=900+x^{2}$$

Combine 40x and 60x to get 100x.

$$100x=900+x^{2}$$

Subtract x^{2} from both sides.

$$100x-x^{2}=900$$

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.

$$-x^{2}+100x=900$$

Divide both sides by -1.

$$\frac{-x^{2}+100x}{-1}=\frac{900}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{100}{-1}x=\frac{900}{-1}$$

Divide 100 by -1.

$$x^{2}-100x=\frac{900}{-1}$$

Divide 900 by -1.

$$x^{2}-100x=-900$$

Divide -100, the coefficient of the x term, by 2 to get -50. Then add the square of -50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-100x+\left(-50\right)^{2}=-900+\left(-50\right)^{2}$$

Square -50.

$$x^{2}-100x+2500=-900+2500$$

Add -900 to 2500.

$$x^{2}-100x+2500=1600$$

Factor x^{2}-100x+2500. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-50\right)^{2}=1600$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-50\right)^{2}}=\sqrt{1600}$$

Simplify.

$$x-50=40 x-50=-40$$

Add 50 to both sides of the equation.

$$x=90 x=10$$