Solve 80/x+4+3x=26

$$\frac { 80 } { x + 4 } + 3 x = 26$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$80+3x\left(x+4\right)=26\left(x+4\right)$$

Use the distributive property to multiply 3x by x+4.

$$80+3x^{2}+12x=26\left(x+4\right)$$

Use the distributive property to multiply 26 by x+4.

$$80+3x^{2}+12x=26x+104$$

Subtract 26x from both sides.

$$80+3x^{2}+12x-26x=104$$

Combine 12x and -26x to get -14x.

$$80+3x^{2}-14x=104$$

Subtract 104 from both sides.

$$80+3x^{2}-14x-104=0$$

Subtract 104 from 80 to get -24.

$$-24+3x^{2}-14x=0$$

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.

$$3x^{2}-14x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -14 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-14\right)±\sqrt{196-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$x=\frac{-\left(-14\right)±\sqrt{196+288}}{2\times 3}$$

Add 196 to 288.

$$x=\frac{-\left(-14\right)±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$x=\frac{-\left(-14\right)±22}{2\times 3}$$

The opposite of -14 is 14.

$$x=\frac{14±22}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{14±22}{6}$$

Now solve the equation x=\frac{14±22}{6} when ± is plus. Add 14 to 22.

$$x=\frac{36}{6}$$

Divide 36 by 6.

$$x=6$$

Now solve the equation x=\frac{14±22}{6} when ± is minus. Subtract 22 from 14.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=6 x=-\frac{4}{3}$$

Steps for Completing the Square

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$80+3x\left(x+4\right)=26\left(x+4\right)$$

Use the distributive property to multiply 3x by x+4.

$$80+3x^{2}+12x=26\left(x+4\right)$$

Use the distributive property to multiply 26 by x+4.

$$80+3x^{2}+12x=26x+104$$

Subtract 26x from both sides.

$$80+3x^{2}+12x-26x=104$$

Combine 12x and -26x to get -14x.

$$80+3x^{2}-14x=104$$

Subtract 80 from both sides.

$$3x^{2}-14x=104-80$$

Subtract 80 from 104 to get 24.

$$3x^{2}-14x=24$$

Divide both sides by 3.

$$\frac{3x^{2}-14x}{3}=\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{14}{3}x=\frac{24}{3}$$

Divide 24 by 3.

$$x^{2}-\frac{14}{3}x=8$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=8+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=8+\frac{49}{9}$$

Add 8 to \frac{49}{9}.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{121}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{11}{3} x-\frac{7}{3}=-\frac{11}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$x=6 x=-\frac{4}{3}$$