Solve x^2+5/3+x/2=4

$$\frac { x ^ { 2 } + 5 } { 3 } + \frac { x } { 2 } = 4$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=2$$

Steps Using Factoring By Grouping

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

$$2\left(x^{2}+5\right)+3x=24$$

Use the distributive property to multiply 2 by x^{2}+5.

$$2x^{2}+10+3x=24$$

Subtract 24 from both sides.

$$2x^{2}+10+3x-24=0$$

Subtract 24 from 10 to get -14.

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

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

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

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

$$a+b=3 ab=2\left(-14\right)=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite 2x^{2}+3x-14 as \left(2x^{2}-4x\right)+\left(7x-14\right).

$$\left(2x^{2}-4x\right)+\left(7x-14\right)$$

Factor out 2x in the first and 7 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$2\left(x^{2}+5\right)+3x=24$$

Use the distributive property to multiply 2 by x^{2}+5.

$$2x^{2}+10+3x=24$$

Subtract 24 from both sides.

$$2x^{2}+10+3x-24=0$$

Subtract 24 from 10 to get -14.

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

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.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-14\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-14\right)}}{2\times 2}$$

Multiply -8 times -14.

$$x=\frac{-3±\sqrt{9+112}}{2\times 2}$$

Add 9 to 112.

$$x=\frac{-3±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-3±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±11}{4}$$

Now solve the equation x=\frac{-3±11}{4} when ± is plus. Add -3 to 11.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-3±11}{4} when ± is minus. Subtract 11 from -3.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$2\left(x^{2}+5\right)+3x=24$$

Use the distributive property to multiply 2 by x^{2}+5.

$$2x^{2}+10+3x=24$$

Subtract 10 from both sides.

$$2x^{2}+3x=24-10$$

Subtract 10 from 24 to get 14.

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

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{14}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{14}{2}$$

Divide 14 by 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=7+\frac{9}{16}$$

Add 7 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{121}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{11}{4} x+\frac{3}{4}=-\frac{11}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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