Solve 2x^2+5=7x

$$2 { x }^{ 2 } +5=7x$$

Solve for x

$$x=1$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$2x^{2}+5-7x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}-7x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 5=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite 2x^{2}-7x+5 as \left(2x^{2}-5x\right)+\left(-2x+5\right).

$$\left(2x^{2}-5x\right)+\left(-2x+5\right)$$

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

$$x\left(2x-5\right)-\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x-1\right)$$

To find equation solutions, solve 2x-5=0 and x-1=0.

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

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$2x^{2}+5-7x=0$$

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.

$$2x^{2}-7x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 5}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-\left(-7\right)±\sqrt{49-40}}{2\times 2}$$

Add 49 to -40.

$$x=\frac{-\left(-7\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-\left(-7\right)±3}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±3}{4}$$

Now solve the equation x=\frac{7±3}{4} when ± is plus. Add 7 to 3.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{7±3}{4} when ± is minus. Subtract 3 from 7.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 7x from both sides.

$$2x^{2}+5-7x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-7x=-5$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=-\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=-\frac{5}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{5}{2}+\frac{49}{16}$$

Add -\frac{5}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{3}{4} x-\frac{7}{4}=-\frac{3}{4}$$

Add \frac{7}{4} to both sides of the equation.

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