Solve (-50+6x)(-50+16x)=left(-50+14xright)^2

$$(-50+6x)(-50+16x)= { \left(-50+14x \right) }^{ 2 }$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply -50+6x by -50+16x and combine like terms.

$$2500-1100x+96x^{2}=\left(-50+14x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-50+14x\right)^{2}.

$$2500-1100x+96x^{2}=2500-1400x+196x^{2}$$

Subtract 2500 from both sides.

$$2500-1100x+96x^{2}-2500=-1400x+196x^{2}$$

Subtract 2500 from 2500 to get 0.

$$-1100x+96x^{2}=-1400x+196x^{2}$$

Add 1400x to both sides.

$$-1100x+96x^{2}+1400x=196x^{2}$$

Combine -1100x and 1400x to get 300x.

$$300x+96x^{2}=196x^{2}$$

Subtract 196x^{2} from both sides.

$$300x+96x^{2}-196x^{2}=0$$

Combine 96x^{2} and -196x^{2} to get -100x^{2}.

$$300x-100x^{2}=0$$

Factor out x.

$$x\left(300-100x\right)=0$$

To find equation solutions, solve x=0 and 300-100x=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -50+6x by -50+16x and combine like terms.

$$2500-1100x+96x^{2}=\left(-50+14x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-50+14x\right)^{2}.

$$2500-1100x+96x^{2}=2500-1400x+196x^{2}$$

Subtract 2500 from both sides.

$$2500-1100x+96x^{2}-2500=-1400x+196x^{2}$$

Subtract 2500 from 2500 to get 0.

$$-1100x+96x^{2}=-1400x+196x^{2}$$

Add 1400x to both sides.

$$-1100x+96x^{2}+1400x=196x^{2}$$

Combine -1100x and 1400x to get 300x.

$$300x+96x^{2}=196x^{2}$$

Subtract 196x^{2} from both sides.

$$300x+96x^{2}-196x^{2}=0$$

Combine 96x^{2} and -196x^{2} to get -100x^{2}.

$$300x-100x^{2}=0$$

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.

$$-100x^{2}+300x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -100 for a, 300 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-300±\sqrt{300^{2}}}{2\left(-100\right)}$$

Take the square root of 300^{2}.

$$x=\frac{-300±300}{2\left(-100\right)}$$

Multiply 2 times -100.

$$x=\frac{-300±300}{-200}$$

Now solve the equation x=\frac{-300±300}{-200} when ± is plus. Add -300 to 300.

$$x=\frac{0}{-200}$$

Divide 0 by -200.

$$x=0$$

Now solve the equation x=\frac{-300±300}{-200} when ± is minus. Subtract 300 from -300.

$$x=-\frac{600}{-200}$$

Divide -600 by -200.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

Use the distributive property to multiply -50+6x by -50+16x and combine like terms.

$$2500-1100x+96x^{2}=\left(-50+14x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-50+14x\right)^{2}.

$$2500-1100x+96x^{2}=2500-1400x+196x^{2}$$

Add 1400x to both sides.

$$2500-1100x+96x^{2}+1400x=2500+196x^{2}$$

Combine -1100x and 1400x to get 300x.

$$2500+300x+96x^{2}=2500+196x^{2}$$

Subtract 196x^{2} from both sides.

$$2500+300x+96x^{2}-196x^{2}=2500$$

Combine 96x^{2} and -196x^{2} to get -100x^{2}.

$$2500+300x-100x^{2}=2500$$

Subtract 2500 from both sides.

$$300x-100x^{2}=2500-2500$$

Subtract 2500 from 2500 to get 0.

$$300x-100x^{2}=0$$

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.

$$-100x^{2}+300x=0$$

Divide both sides by -100.

$$\frac{-100x^{2}+300x}{-100}=\frac{0}{-100}$$

Dividing by -100 undoes the multiplication by -100.

$$x^{2}+\frac{300}{-100}x=\frac{0}{-100}$$

Divide 300 by -100.

$$x^{2}-3x=\frac{0}{-100}$$

Divide 0 by -100.

$$x^{2}-3x=0$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$