Solve (3)4(x-3)^2-25(x-2)^2=0

$$( 3 ) 4 ( x - 3 ) ^ { 2 } - 25 ( x - 2 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{10 \sqrt{3} + 14}{13} \approx 2.409269852$$

$$x=\frac{14-10\sqrt{3}}{13}\approx -0.255423698$$

Steps Using the Quadratic Formula

Multiply 3 and 4 to get 12.

$$12\left(x-3\right)^{2}-25\left(x-2\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$12\left(x^{2}-6x+9\right)-25\left(x-2\right)^{2}=0$$

Use the distributive property to multiply 12 by x^{2}-6x+9.

$$12x^{2}-72x+108-25\left(x-2\right)^{2}=0$$

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

$$12x^{2}-72x+108-25\left(x^{2}-4x+4\right)=0$$

Use the distributive property to multiply -25 by x^{2}-4x+4.

$$12x^{2}-72x+108-25x^{2}+100x-100=0$$

Combine 12x^{2} and -25x^{2} to get -13x^{2}.

$$-13x^{2}-72x+108+100x-100=0$$

Combine -72x and 100x to get 28x.

$$-13x^{2}+28x+108-100=0$$

Subtract 100 from 108 to get 8.

$$-13x^{2}+28x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -13 for a, 28 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\left(-13\right)\times 8}}{2\left(-13\right)}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\left(-13\right)\times 8}}{2\left(-13\right)}$$

Multiply -4 times -13.

$$x=\frac{-28±\sqrt{784+52\times 8}}{2\left(-13\right)}$$

Multiply 52 times 8.

$$x=\frac{-28±\sqrt{784+416}}{2\left(-13\right)}$$

Add 784 to 416.

$$x=\frac{-28±\sqrt{1200}}{2\left(-13\right)}$$

Take the square root of 1200.

$$x=\frac{-28±20\sqrt{3}}{2\left(-13\right)}$$

Multiply 2 times -13.

$$x=\frac{-28±20\sqrt{3}}{-26}$$

Now solve the equation x=\frac{-28±20\sqrt{3}}{-26} when ± is plus. Add -28 to 20\sqrt{3}.

$$x=\frac{20\sqrt{3}-28}{-26}$$

Divide -28+20\sqrt{3} by -26.

$$x=\frac{14-10\sqrt{3}}{13}$$

Now solve the equation x=\frac{-28±20\sqrt{3}}{-26} when ± is minus. Subtract 20\sqrt{3} from -28.

$$x=\frac{-20\sqrt{3}-28}{-26}$$

Divide -28-20\sqrt{3} by -26.

$$x=\frac{10\sqrt{3}+14}{13}$$

The equation is now solved.

$$x=\frac{14-10\sqrt{3}}{13} x=\frac{10\sqrt{3}+14}{13}$$

Steps for Completing the Square

Multiply 3 and 4 to get 12.

$$12\left(x-3\right)^{2}-25\left(x-2\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$12\left(x^{2}-6x+9\right)-25\left(x-2\right)^{2}=0$$

Use the distributive property to multiply 12 by x^{2}-6x+9.

$$12x^{2}-72x+108-25\left(x-2\right)^{2}=0$$

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

$$12x^{2}-72x+108-25\left(x^{2}-4x+4\right)=0$$

Use the distributive property to multiply -25 by x^{2}-4x+4.

$$12x^{2}-72x+108-25x^{2}+100x-100=0$$

Combine 12x^{2} and -25x^{2} to get -13x^{2}.

$$-13x^{2}-72x+108+100x-100=0$$

Combine -72x and 100x to get 28x.

$$-13x^{2}+28x+108-100=0$$

Subtract 100 from 108 to get 8.

$$-13x^{2}+28x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-13x^{2}+28x=-8$$

Divide both sides by -13.

$$\frac{-13x^{2}+28x}{-13}=-\frac{8}{-13}$$

Dividing by -13 undoes the multiplication by -13.

$$x^{2}+\frac{28}{-13}x=-\frac{8}{-13}$$

Divide 28 by -13.

$$x^{2}-\frac{28}{13}x=-\frac{8}{-13}$$

Divide -8 by -13.

$$x^{2}-\frac{28}{13}x=\frac{8}{13}$$

Divide -\frac{28}{13}, the coefficient of the x term, by 2 to get -\frac{14}{13}. Then add the square of -\frac{14}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{28}{13}x+\left(-\frac{14}{13}\right)^{2}=\frac{8}{13}+\left(-\frac{14}{13}\right)^{2}$$

Square -\frac{14}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{28}{13}x+\frac{196}{169}=\frac{8}{13}+\frac{196}{169}$$

Add \frac{8}{13} to \frac{196}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{28}{13}x+\frac{196}{169}=\frac{300}{169}$$

Factor x^{2}-\frac{28}{13}x+\frac{196}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{14}{13}\right)^{2}=\frac{300}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{14}{13}\right)^{2}}=\sqrt{\frac{300}{169}}$$

Simplify.

$$x-\frac{14}{13}=\frac{10\sqrt{3}}{13} x-\frac{14}{13}=-\frac{10\sqrt{3}}{13}$$

Add \frac{14}{13} to both sides of the equation.

$$x=\frac{10\sqrt{3}+14}{13} x=\frac{14-10\sqrt{3}}{13}$$