Solve 6x^2=11x+7

$$6 x ^ { 2 } = 11 x + 7$$

Solve for x

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

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Subtract 11x from both sides.

$$6x^{2}-11x=7$$

Subtract 7 from both sides.

$$6x^{2}-11x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-11 ab=6\left(-7\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -11.

$$a=-14 b=3$$

Rewrite 6x^{2}-11x-7 as \left(6x^{2}-14x\right)+\left(3x-7\right).

$$\left(6x^{2}-14x\right)+\left(3x-7\right)$$

Factor out 2x in 6x^{2}-14x.

$$2x\left(3x-7\right)+3x-7$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(2x+1\right)$$

To find equation solutions, solve 3x-7=0 and 2x+1=0.

$$x=\frac{7}{3} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$6x^{2}-11x=7$$

Subtract 7 from both sides.

$$6x^{2}-11x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -11 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-7\right)}}{2\times 6}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-7\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-24\left(-7\right)}}{2\times 6}$$

Multiply -24 times -7.

$$x=\frac{-\left(-11\right)±\sqrt{121+168}}{2\times 6}$$

Add 121 to 168.

$$x=\frac{-\left(-11\right)±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$x=\frac{-\left(-11\right)±17}{2\times 6}$$

The opposite of -11 is 11.

$$x=\frac{11±17}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{11±17}{12}$$

Now solve the equation x=\frac{11±17}{12} when ± is plus. Add 11 to 17.

$$x=\frac{28}{12}$$

Reduce the fraction \frac{28}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{11±17}{12} when ± is minus. Subtract 17 from 11.

$$x=-\frac{6}{12}$$

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=\frac{7}{3} x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 11x from both sides.

$$6x^{2}-11x=7$$

Divide both sides by 6.

$$\frac{6x^{2}-11x}{6}=\frac{7}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{11}{6}x=\frac{7}{6}$$

Divide -\frac{11}{6}, the coefficient of the x term, by 2 to get -\frac{11}{12}. Then add the square of -\frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{6}x+\left(-\frac{11}{12}\right)^{2}=\frac{7}{6}+\left(-\frac{11}{12}\right)^{2}$$

Square -\frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{6}x+\frac{121}{144}=\frac{7}{6}+\frac{121}{144}$$

Add \frac{7}{6} to \frac{121}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{6}x+\frac{121}{144}=\frac{289}{144}$$

Factor x^{2}-\frac{11}{6}x+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$x-\frac{11}{12}=\frac{17}{12} x-\frac{11}{12}=-\frac{17}{12}$$

Add \frac{11}{12} to both sides of the equation.

$$x=\frac{7}{3} x=-\frac{1}{2}$$