Solve x^2+x/2=8x-7/3

$$\frac{ { x }^{ 2 } +x }{ 2 } = \frac{ 8x-7 }{ 3 }$$

Solve for x

$$x=2$$

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

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(x^{2}+x\right)=2\left(8x-7\right)$$

Use the distributive property to multiply 3 by x^{2}+x.

$$3x^{2}+3x=2\left(8x-7\right)$$

Use the distributive property to multiply 2 by 8x-7.

$$3x^{2}+3x=16x-14$$

Subtract 16x from both sides.

$$3x^{2}+3x-16x=-14$$

Combine 3x and -16x to get -13x.

$$3x^{2}-13x=-14$$

Add 14 to both sides.

$$3x^{2}-13x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-13 ab=3\times 14=42$$

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 42.

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

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite 3x^{2}-13x+14 as \left(3x^{2}-7x\right)+\left(-6x+14\right).

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(x^{2}+x\right)=2\left(8x-7\right)$$

Use the distributive property to multiply 3 by x^{2}+x.

$$3x^{2}+3x=2\left(8x-7\right)$$

Use the distributive property to multiply 2 by 8x-7.

$$3x^{2}+3x=16x-14$$

Subtract 16x from both sides.

$$3x^{2}+3x-16x=-14$$

Combine 3x and -16x to get -13x.

$$3x^{2}-13x=-14$$

Add 14 to both sides.

$$3x^{2}-13x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -13 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\times 14}}{2\times 3}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 3\times 14}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-13\right)±\sqrt{169-12\times 14}}{2\times 3}$$

Multiply -12 times 14.

$$x=\frac{-\left(-13\right)±\sqrt{169-168}}{2\times 3}$$

Add 169 to -168.

$$x=\frac{-\left(-13\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-\left(-13\right)±1}{2\times 3}$$

The opposite of -13 is 13.

$$x=\frac{13±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{13±1}{6}$$

Now solve the equation x=\frac{13±1}{6} when ± is plus. Add 13 to 1.

$$x=\frac{14}{6}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{13±1}{6} when ± is minus. Subtract 1 from 13.

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

Divide 12 by 6.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(x^{2}+x\right)=2\left(8x-7\right)$$

Use the distributive property to multiply 3 by x^{2}+x.

$$3x^{2}+3x=2\left(8x-7\right)$$

Use the distributive property to multiply 2 by 8x-7.

$$3x^{2}+3x=16x-14$$

Subtract 16x from both sides.

$$3x^{2}+3x-16x=-14$$

Combine 3x and -16x to get -13x.

$$3x^{2}-13x=-14$$

Divide both sides by 3.

$$\frac{3x^{2}-13x}{3}=-\frac{14}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{13}{3}x=-\frac{14}{3}$$

Divide -\frac{13}{3}, the coefficient of the x term, by 2 to get -\frac{13}{6}. Then add the square of -\frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{3}x+\left(-\frac{13}{6}\right)^{2}=-\frac{14}{3}+\left(-\frac{13}{6}\right)^{2}$$

Square -\frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{3}x+\frac{169}{36}=-\frac{14}{3}+\frac{169}{36}$$

Add -\frac{14}{3} to \frac{169}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{3}x+\frac{169}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{13}{3}x+\frac{169}{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}.

$$\left(x-\frac{13}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{13}{6}=\frac{1}{6} x-\frac{13}{6}=-\frac{1}{6}$$

Add \frac{13}{6} to both sides of the equation.

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