Solve -128(1+x)+128(1+x)^2=608

$$- 128 ( 1 + x ) + 128 ( 1 + x ) ^ { 2 } = 608$$

Solve for x

$$x=\sqrt{5}-\frac{1}{2}\approx 1.736067977$$

$$x=-\sqrt{5}-\frac{1}{2}\approx -2.736067977$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -128 by 1+x.

$$-128-128x+128\left(1+x\right)^{2}=608$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$-128-128x+128\left(1+2x+x^{2}\right)=608$$

Use the distributive property to multiply 128 by 1+2x+x^{2}.

$$-128-128x+128+256x+128x^{2}=608$$

Add -128 and 128 to get 0.

$$-128x+256x+128x^{2}=608$$

Combine -128x and 256x to get 128x.

$$128x+128x^{2}=608$$

Subtract 608 from both sides.

$$128x+128x^{2}-608=0$$

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.

$$128x^{2}+128x-608=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 128 for a, 128 for b, and -608 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-128±\sqrt{128^{2}-4\times 128\left(-608\right)}}{2\times 128}$$

Square 128.

$$x=\frac{-128±\sqrt{16384-4\times 128\left(-608\right)}}{2\times 128}$$

Multiply -4 times 128.

$$x=\frac{-128±\sqrt{16384-512\left(-608\right)}}{2\times 128}$$

Multiply -512 times -608.

$$x=\frac{-128±\sqrt{16384+311296}}{2\times 128}$$

Add 16384 to 311296.

$$x=\frac{-128±\sqrt{327680}}{2\times 128}$$

Take the square root of 327680.

$$x=\frac{-128±256\sqrt{5}}{2\times 128}$$

Multiply 2 times 128.

$$x=\frac{-128±256\sqrt{5}}{256}$$

Now solve the equation x=\frac{-128±256\sqrt{5}}{256} when ± is plus. Add -128 to 256\sqrt{5}.

$$x=\frac{256\sqrt{5}-128}{256}$$

Divide -128+256\sqrt{5} by 256.

$$x=\sqrt{5}-\frac{1}{2}$$

Now solve the equation x=\frac{-128±256\sqrt{5}}{256} when ± is minus. Subtract 256\sqrt{5} from -128.

$$x=\frac{-256\sqrt{5}-128}{256}$$

Divide -128-256\sqrt{5} by 256.

$$x=-\sqrt{5}-\frac{1}{2}$$

The equation is now solved.

$$x=\sqrt{5}-\frac{1}{2} x=-\sqrt{5}-\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply -128 by 1+x.

$$-128-128x+128\left(1+x\right)^{2}=608$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$-128-128x+128\left(1+2x+x^{2}\right)=608$$

Use the distributive property to multiply 128 by 1+2x+x^{2}.

$$-128-128x+128+256x+128x^{2}=608$$

Add -128 and 128 to get 0.

$$-128x+256x+128x^{2}=608$$

Combine -128x and 256x to get 128x.

$$128x+128x^{2}=608$$

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.

$$128x^{2}+128x=608$$

Divide both sides by 128.

$$\frac{128x^{2}+128x}{128}=\frac{608}{128}$$

Dividing by 128 undoes the multiplication by 128.

$$x^{2}+\frac{128}{128}x=\frac{608}{128}$$

Divide 128 by 128.

$$x^{2}+x=\frac{608}{128}$$

Reduce the fraction \frac{608}{128} to lowest terms by extracting and canceling out 32.

$$x^{2}+x=\frac{19}{4}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{19}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{19+1}{4}$$

Add \frac{19}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=5$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+\frac{1}{2}=\sqrt{5} x+\frac{1}{2}=-\sqrt{5}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\sqrt{5}-\frac{1}{2} x=-\sqrt{5}-\frac{1}{2}$$