Solve 2x^2-5x+2=0quadt

$$2 x ^ { 2 } - 5 x + 2 = 0 \quad t$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

Anything times zero gives zero.

$$2x^{2}-5x+2=0$$

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

$$a+b=-5 ab=2\times 2=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite 2x^{2}-5x+2 as \left(2x^{2}-4x\right)+\left(-x+2\right).

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

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

$$2x\left(x-2\right)-\left(x-2\right)$$

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

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

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

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

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$2x^{2}-5x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 2}$$

Add 25 to -16.

$$x=\frac{-\left(-5\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

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

The opposite of -5 is 5.

$$x=\frac{5±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±3}{4}$$

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

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

Divide 8 by 4.

$$x=2$$

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

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Anything times zero gives zero.

$$2x^{2}-5x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-5x=-2$$

Divide both sides by 2.

$$\frac{2x^{2}-5x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{5}{2}x=-\frac{2}{2}$$

Divide -2 by 2.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-1+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Add -1 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{3}{4} x-\frac{5}{4}=-\frac{3}{4}$$

Add \frac{5}{4} to both sides of the equation.

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

Solve for t (complex solution)

$$t\in \mathrm{C}$$

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

Solve for t

$$t\in \mathrm{R}$$

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