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

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

Solve for x

$$x=\sqrt{6}+6\approx 8.449489743$$

$$x=6-\sqrt{6}\approx 3.550510257$$

Steps Using the Quadratic Formula

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

$$2\left(x^{2}-6x+9\right)=3\left(x-4\right)^{2}$$

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

$$2x^{2}-12x+18=3\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x^{2}-12x+18=3\left(x^{2}-8x+16\right)$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$2x^{2}-12x+18=3x^{2}-24x+48$$

Subtract 3x^{2} from both sides.

$$2x^{2}-12x+18-3x^{2}=-24x+48$$

Combine 2x^{2} and -3x^{2} to get -x^{2}.

$$-x^{2}-12x+18=-24x+48$$

Add 24x to both sides.

$$-x^{2}-12x+18+24x=48$$

Combine -12x and 24x to get 12x.

$$-x^{2}+12x+18=48$$

Subtract 48 from both sides.

$$-x^{2}+12x+18-48=0$$

Subtract 48 from 18 to get -30.

$$-x^{2}+12x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 12 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-12±\sqrt{144+4\left(-30\right)}}{2\left(-1\right)}$$

Multiply 4 times -30.

$$x=\frac{-12±\sqrt{144-120}}{2\left(-1\right)}$$

Add 144 to -120.

$$x=\frac{-12±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-12±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-12±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{-12±2\sqrt{6}}{-2} when ± is plus. Add -12 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-12}{-2}$$

Divide -12+2\sqrt{6} by -2.

$$x=6-\sqrt{6}$$

Now solve the equation x=\frac{-12±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from -12.

$$x=\frac{-2\sqrt{6}-12}{-2}$$

Divide -12-2\sqrt{6} by -2.

$$x=\sqrt{6}+6$$

The equation is now solved.

$$x=6-\sqrt{6} x=\sqrt{6}+6$$

Steps for Completing the Square

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

$$2\left(x^{2}-6x+9\right)=3\left(x-4\right)^{2}$$

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

$$2x^{2}-12x+18=3\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x^{2}-12x+18=3\left(x^{2}-8x+16\right)$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$2x^{2}-12x+18=3x^{2}-24x+48$$

Subtract 3x^{2} from both sides.

$$2x^{2}-12x+18-3x^{2}=-24x+48$$

Combine 2x^{2} and -3x^{2} to get -x^{2}.

$$-x^{2}-12x+18=-24x+48$$

Add 24x to both sides.

$$-x^{2}-12x+18+24x=48$$

Combine -12x and 24x to get 12x.

$$-x^{2}+12x+18=48$$

Subtract 18 from both sides.

$$-x^{2}+12x=48-18$$

Subtract 18 from 48 to get 30.

$$-x^{2}+12x=30$$

Divide both sides by -1.

$$\frac{-x^{2}+12x}{-1}=\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{12}{-1}x=\frac{30}{-1}$$

Divide 12 by -1.

$$x^{2}-12x=\frac{30}{-1}$$

Divide 30 by -1.

$$x^{2}-12x=-30$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-30+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-30+36$$

Add -30 to 36.

$$x^{2}-12x+36=6$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-6=\sqrt{6} x-6=-\sqrt{6}$$

Add 6 to both sides of the equation.

$$x=\sqrt{6}+6 x=6-\sqrt{6}$$