Solve 2x^2+6=7x

$$2 x ^ { 2 } + 6 = 7 x$$

Solve for x

$$x=2$$

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

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$2x^{2}+6-7x=0$$

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

$$2x^{2}-7x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 6=12$$

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 -7.

$$a=-4 b=-3$$

Rewrite 2x^{2}-7x+6 as \left(2x^{2}-4x\right)+\left(-3x+6\right).

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

Factor out 2x in the first and -3 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$2x^{2}+6-7x=0$$

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.

$$2x^{2}-7x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 2}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±1}{4}$$

Now solve the equation x=\frac{7±1}{4} when ± is plus. Add 7 to 1.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{7±1}{4} when ± is minus. Subtract 1 from 7.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 7x from both sides.

$$2x^{2}+6-7x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-7x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=-\frac{6}{2}$$

Divide -6 by 2.

$$x^{2}-\frac{7}{2}x=-3$$

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.

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

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

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-3+\frac{49}{16}$$

Add -3 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{1}{16}$$

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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{1}{4} x-\frac{7}{4}=-\frac{1}{4}$$

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

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