Solve -2.4(x-2)-(x-2)^2=1/2(x-0.4)^2

$$- 2.4 ( x - 2 ) - ( x - 2 ) ^ { 2 } = \frac { 1 } { 2 } ( x - 0.4 ) ^ { 2 }$$

Solve for x

$$x=\frac{4\sqrt{13}}{15}+\frac{2}{3}\approx 1.628147007$$

$$x=-\frac{4\sqrt{13}}{15}+\frac{2}{3}\approx -0.294813673$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -2.4 by x-2.

$$-2.4x+4.8-\left(x-2\right)^{2}=\frac{1}{2}\left(x-0.4\right)^{2}$$

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

$$-2.4x+4.8-\left(x^{2}-4x+4\right)=\frac{1}{2}\left(x-0.4\right)^{2}$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$-2.4x+4.8-x^{2}+4x-4=\frac{1}{2}\left(x-0.4\right)^{2}$$

Combine -2.4x and 4x to get 1.6x.

$$1.6x+4.8-x^{2}-4=\frac{1}{2}\left(x-0.4\right)^{2}$$

Subtract 4 from 4.8 to get 0.8.

$$1.6x+0.8-x^{2}=\frac{1}{2}\left(x-0.4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-0.4\right)^{2}.

$$1.6x+0.8-x^{2}=\frac{1}{2}\left(x^{2}-0.8x+0.16\right)$$

Use the distributive property to multiply \frac{1}{2} by x^{2}-0.8x+0.16.

$$1.6x+0.8-x^{2}=\frac{1}{2}x^{2}-\frac{2}{5}x+\frac{2}{25}$$

Subtract \frac{1}{2}x^{2} from both sides.

$$1.6x+0.8-x^{2}-\frac{1}{2}x^{2}=-\frac{2}{5}x+\frac{2}{25}$$

Combine -x^{2} and -\frac{1}{2}x^{2} to get -\frac{3}{2}x^{2}.

$$1.6x+0.8-\frac{3}{2}x^{2}=-\frac{2}{5}x+\frac{2}{25}$$

Add \frac{2}{5}x to both sides.

$$1.6x+0.8-\frac{3}{2}x^{2}+\frac{2}{5}x=\frac{2}{25}$$

Combine 1.6x and \frac{2}{5}x to get 2x.

$$2x+0.8-\frac{3}{2}x^{2}=\frac{2}{25}$$

Subtract \frac{2}{25} from both sides.

$$2x+0.8-\frac{3}{2}x^{2}-\frac{2}{25}=0$$

Subtract \frac{2}{25} from 0.8 to get \frac{18}{25}.

$$2x+\frac{18}{25}-\frac{3}{2}x^{2}=0$$

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.

$$-\frac{3}{2}x^{2}+2x+\frac{18}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{2} for a, 2 for b, and \frac{18}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-\frac{3}{2}\right)\times \frac{18}{25}}}{2\left(-\frac{3}{2}\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{3}{2}\right)\times \frac{18}{25}}}{2\left(-\frac{3}{2}\right)}$$

Multiply -4 times -\frac{3}{2}.

$$x=\frac{-2±\sqrt{4+6\times \frac{18}{25}}}{2\left(-\frac{3}{2}\right)}$$

Multiply 6 times \frac{18}{25}.

$$x=\frac{-2±\sqrt{4+\frac{108}{25}}}{2\left(-\frac{3}{2}\right)}$$

Add 4 to \frac{108}{25}.

$$x=\frac{-2±\sqrt{\frac{208}{25}}}{2\left(-\frac{3}{2}\right)}$$

Take the square root of \frac{208}{25}.

$$x=\frac{-2±\frac{4\sqrt{13}}{5}}{2\left(-\frac{3}{2}\right)}$$

Multiply 2 times -\frac{3}{2}.

$$x=\frac{-2±\frac{4\sqrt{13}}{5}}{-3}$$

Now solve the equation x=\frac{-2±\frac{4\sqrt{13}}{5}}{-3} when ± is plus. Add -2 to \frac{4\sqrt{13}}{5}.

$$x=\frac{\frac{4\sqrt{13}}{5}-2}{-3}$$

Divide -2+\frac{4\sqrt{13}}{5} by -3.

$$x=-\frac{4\sqrt{13}}{15}+\frac{2}{3}$$

Now solve the equation x=\frac{-2±\frac{4\sqrt{13}}{5}}{-3} when ± is minus. Subtract \frac{4\sqrt{13}}{5} from -2.

$$x=\frac{-\frac{4\sqrt{13}}{5}-2}{-3}$$

Divide -2-\frac{4\sqrt{13}}{5} by -3.

$$x=\frac{4\sqrt{13}}{15}+\frac{2}{3}$$

The equation is now solved.

$$x=-\frac{4\sqrt{13}}{15}+\frac{2}{3} x=\frac{4\sqrt{13}}{15}+\frac{2}{3}$$

Steps for Completing the Square

Use the distributive property to multiply -2.4 by x-2.

$$-2.4x+4.8-\left(x-2\right)^{2}=\frac{1}{2}\left(x-0.4\right)^{2}$$

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

$$-2.4x+4.8-\left(x^{2}-4x+4\right)=\frac{1}{2}\left(x-0.4\right)^{2}$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$-2.4x+4.8-x^{2}+4x-4=\frac{1}{2}\left(x-0.4\right)^{2}$$

Combine -2.4x and 4x to get 1.6x.

$$1.6x+4.8-x^{2}-4=\frac{1}{2}\left(x-0.4\right)^{2}$$

Subtract 4 from 4.8 to get 0.8.

$$1.6x+0.8-x^{2}=\frac{1}{2}\left(x-0.4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-0.4\right)^{2}.

$$1.6x+0.8-x^{2}=\frac{1}{2}\left(x^{2}-0.8x+0.16\right)$$

Use the distributive property to multiply \frac{1}{2} by x^{2}-0.8x+0.16.

$$1.6x+0.8-x^{2}=\frac{1}{2}x^{2}-\frac{2}{5}x+\frac{2}{25}$$

Subtract \frac{1}{2}x^{2} from both sides.

$$1.6x+0.8-x^{2}-\frac{1}{2}x^{2}=-\frac{2}{5}x+\frac{2}{25}$$

Combine -x^{2} and -\frac{1}{2}x^{2} to get -\frac{3}{2}x^{2}.

$$1.6x+0.8-\frac{3}{2}x^{2}=-\frac{2}{5}x+\frac{2}{25}$$

Add \frac{2}{5}x to both sides.

$$1.6x+0.8-\frac{3}{2}x^{2}+\frac{2}{5}x=\frac{2}{25}$$

Combine 1.6x and \frac{2}{5}x to get 2x.

$$2x+0.8-\frac{3}{2}x^{2}=\frac{2}{25}$$

Subtract 0.8 from both sides.

$$2x-\frac{3}{2}x^{2}=\frac{2}{25}-0.8$$

Subtract 0.8 from \frac{2}{25} to get -\frac{18}{25}.

$$2x-\frac{3}{2}x^{2}=-\frac{18}{25}$$

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.

$$-\frac{3}{2}x^{2}+2x=-\frac{18}{25}$$

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{2}x^{2}+2x}{-\frac{3}{2}}=-\frac{\frac{18}{25}}{-\frac{3}{2}}$$

Dividing by -\frac{3}{2} undoes the multiplication by -\frac{3}{2}.

$$x^{2}+\frac{2}{-\frac{3}{2}}x=-\frac{\frac{18}{25}}{-\frac{3}{2}}$$

Divide 2 by -\frac{3}{2} by multiplying 2 by the reciprocal of -\frac{3}{2}.

$$x^{2}-\frac{4}{3}x=-\frac{\frac{18}{25}}{-\frac{3}{2}}$$

Divide -\frac{18}{25} by -\frac{3}{2} by multiplying -\frac{18}{25} by the reciprocal of -\frac{3}{2}.

$$x^{2}-\frac{4}{3}x=\frac{12}{25}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{12}{25}+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{12}{25}+\frac{4}{9}$$

Add \frac{12}{25} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{208}{225}$$

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

$$\left(x-\frac{2}{3}\right)^{2}=\frac{208}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{208}{225}}$$

Simplify.

$$x-\frac{2}{3}=\frac{4\sqrt{13}}{15} x-\frac{2}{3}=-\frac{4\sqrt{13}}{15}$$

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

$$x=\frac{4\sqrt{13}}{15}+\frac{2}{3} x=-\frac{4\sqrt{13}}{15}+\frac{2}{3}$$