Solve 3x^2+6=11x

$$3 x ^ { 2 } + 6 = 11 x$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 11x from both sides.

$$3x^{2}+6-11x=0$$

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

$$3x^{2}-11x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 6=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite 3x^{2}-11x+6 as \left(3x^{2}-9x\right)+\left(-2x+6\right).

$$\left(3x^{2}-9x\right)+\left(-2x+6\right)$$

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

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

Factor out common term x-3 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$3x^{2}+6-11x=0$$

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.

$$3x^{2}-11x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-72}}{2\times 3}$$

Add 121 to -72.

$$x=\frac{-\left(-11\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-\left(-11\right)±7}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±7}{6}$$

Now solve the equation x=\frac{11±7}{6} when ± is plus. Add 11 to 7.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{11±7}{6} when ± is minus. Subtract 7 from 11.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 11x from both sides.

$$3x^{2}+6-11x=0$$

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

$$3x^{2}-11x=-6$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=-\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=-\frac{6}{3}$$

Divide -6 by 3.

$$x^{2}-\frac{11}{3}x=-2$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-2+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-2+\frac{121}{36}$$

Add -2 to \frac{121}{36}.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{7}{6} x-\frac{11}{6}=-\frac{7}{6}$$

Add \frac{11}{6} to both sides of the equation.

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