Solve 128+128(1+x)+128(1+x)(1+x)=608

$$128+128(1+x)+128(1+x)(1+x)=608$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

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

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

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

Add 128 and 128 to get 256.

$$256+128x+128\left(1+x\right)^{2}=608$$

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

$$256+128x+128\left(1+2x+x^{2}\right)=608$$

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

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

Add 256 and 128 to get 384.

$$384+128x+256x+128x^{2}=608$$

Combine 128x and 256x to get 384x.

$$384+384x+128x^{2}=608$$

Subtract 608 from both sides.

$$384+384x+128x^{2}-608=0$$

Subtract 608 from 384 to get -224.

$$-224+384x+128x^{2}=0$$

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.

$$128x^{2}+384x-224=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 128 for a, 384 for b, and -224 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-384±\sqrt{384^{2}-4\times 128\left(-224\right)}}{2\times 128}$$

Square 384.

$$x=\frac{-384±\sqrt{147456-4\times 128\left(-224\right)}}{2\times 128}$$

Multiply -4 times 128.

$$x=\frac{-384±\sqrt{147456-512\left(-224\right)}}{2\times 128}$$

Multiply -512 times -224.

$$x=\frac{-384±\sqrt{147456+114688}}{2\times 128}$$

Add 147456 to 114688.

$$x=\frac{-384±\sqrt{262144}}{2\times 128}$$

Take the square root of 262144.

$$x=\frac{-384±512}{2\times 128}$$

Multiply 2 times 128.

$$x=\frac{-384±512}{256}$$

Now solve the equation x=\frac{-384±512}{256} when ± is plus. Add -384 to 512.

$$x=\frac{128}{256}$$

Reduce the fraction \frac{128}{256} to lowest terms by extracting and canceling out 128.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-384±512}{256} when ± is minus. Subtract 512 from -384.

$$x=-\frac{896}{256}$$

Reduce the fraction \frac{-896}{256} to lowest terms by extracting and canceling out 128.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{7}{2}$$

Steps for Completing the Square

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

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

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

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

Add 128 and 128 to get 256.

$$256+128x+128\left(1+x\right)^{2}=608$$

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

$$256+128x+128\left(1+2x+x^{2}\right)=608$$

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

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

Add 256 and 128 to get 384.

$$384+128x+256x+128x^{2}=608$$

Combine 128x and 256x to get 384x.

$$384+384x+128x^{2}=608$$

Subtract 384 from both sides.

$$384x+128x^{2}=608-384$$

Subtract 384 from 608 to get 224.

$$384x+128x^{2}=224$$

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.

$$128x^{2}+384x=224$$

Divide both sides by 128.

$$\frac{128x^{2}+384x}{128}=\frac{224}{128}$$

Dividing by 128 undoes the multiplication by 128.

$$x^{2}+\frac{384}{128}x=\frac{224}{128}$$

Divide 384 by 128.

$$x^{2}+3x=\frac{224}{128}$$

Reduce the fraction \frac{224}{128} to lowest terms by extracting and canceling out 32.

$$x^{2}+3x=\frac{7}{4}$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{7}{4}+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=\frac{7+9}{4}$$

Add \frac{7}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{1}{2} x=-\frac{7}{2}$$