Solve 2(2u^2-4u)+3=0

$$2 ( 2 u ^ { 2 } - 4 u ) + 3 = 0$$

Solve for u

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

$$u = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2 by 2u^{2}-4u.

$$4u^{2}-8u+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4u^{2}+au+bu+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=4\times 3=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite 4u^{2}-8u+3 as \left(4u^{2}-6u\right)+\left(-2u+3\right).

$$\left(4u^{2}-6u\right)+\left(-2u+3\right)$$

Factor out 2u in the first and -1 in the second group.

$$2u\left(2u-3\right)-\left(2u-3\right)$$

Factor out common term 2u-3 by using distributive property.

$$\left(2u-3\right)\left(2u-1\right)$$

To find equation solutions, solve 2u-3=0 and 2u-1=0.

$$u=\frac{3}{2} u=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by 2u^{2}-4u.

$$4u^{2}-8u+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Square -8.

$$u=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$u=\frac{-\left(-8\right)±\sqrt{64-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$u=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 4}$$

Add 64 to -48.

$$u=\frac{-\left(-8\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$u=\frac{-\left(-8\right)±4}{2\times 4}$$

The opposite of -8 is 8.

$$u=\frac{8±4}{2\times 4}$$

Multiply 2 times 4.

$$u=\frac{8±4}{8}$$

Now solve the equation u=\frac{8±4}{8} when ± is plus. Add 8 to 4.

$$u=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$u=\frac{3}{2}$$

Now solve the equation u=\frac{8±4}{8} when ± is minus. Subtract 4 from 8.

$$u=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$u=\frac{3}{2} u=\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by 2u^{2}-4u.

$$4u^{2}-8u+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4u^{2}-8u=-3$$

Divide both sides by 4.

$$\frac{4u^{2}-8u}{4}=-\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$u^{2}+\left(-\frac{8}{4}\right)u=-\frac{3}{4}$$

Divide -8 by 4.

$$u^{2}-2u=-\frac{3}{4}$$

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

$$u^{2}-2u+1=-\frac{3}{4}+1$$

Add -\frac{3}{4} to 1.

$$u^{2}-2u+1=\frac{1}{4}$$

Factor u^{2}-2u+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(u-1\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u-1\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$u-1=\frac{1}{2} u-1=-\frac{1}{2}$$

Add 1 to both sides of the equation.

$$u=\frac{3}{2} u=\frac{1}{2}$$