Solve 128x^2+384x+384=608

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

Solve for x

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

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

Steps Using Factoring By Grouping

Subtract 608 from both sides.

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

Subtract 608 from 384 to get -224.

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

Divide both sides by 32.

$$4x^{2}+12x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=12 ab=4\left(-7\right)=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 12.

$$a=-2 b=14$$

Rewrite 4x^{2}+12x-7 as \left(4x^{2}-2x\right)+\left(14x-7\right).

$$\left(4x^{2}-2x\right)+\left(14x-7\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(2x-1\right)+7\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(2x+7\right)$$

To find equation solutions, solve 2x-1=0 and 2x+7=0.

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

Steps Using the Quadratic Formula

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.

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

Subtract 608 from both sides of the equation.

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

Subtracting 608 from itself leaves 0.

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

Subtract 608 from 384.

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

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

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.

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

Subtract 384 from both sides of the equation.

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

Subtracting 384 from itself leaves 0.

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

Subtract 384 from 608.

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

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.

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