Solve 5(2x+1)+3(2x-6)=-(x-7)3x

$$5(2x+1)+3(2x-6)=-(x-7)3x$$

Solve for x

$$x = \frac{\sqrt{181} + 5}{6} \approx 3.075604008$$

$$x=\frac{5-\sqrt{181}}{6}\approx -1.408937341$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5 by 2x+1.

$$10x+5+3\left(2x-6\right)=\left(-\left(x-7\right)\right)\times 3x$$

Use the distributive property to multiply 3 by 2x-6.

$$10x+5+6x-18=\left(-\left(x-7\right)\right)\times 3x$$

Combine 10x and 6x to get 16x.

$$16x+5-18=\left(-\left(x-7\right)\right)\times 3x$$

Subtract 18 from 5 to get -13.

$$16x-13=\left(-\left(x-7\right)\right)\times 3x$$

To find the opposite of x-7, find the opposite of each term.

$$16x-13=\left(-x-\left(-7\right)\right)\times 3x$$

The opposite of -7 is 7.

$$16x-13=\left(-x+7\right)\times 3x$$

Use the distributive property to multiply -x+7 by 3.

$$16x-13=\left(-3x+21\right)x$$

Use the distributive property to multiply -3x+21 by x.

$$16x-13=-3x^{2}+21x$$

Add 3x^{2} to both sides.

$$16x-13+3x^{2}=21x$$

Subtract 21x from both sides.

$$16x-13+3x^{2}-21x=0$$

Combine 16x and -21x to get -5x.

$$-5x-13+3x^{2}=0$$

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.

$$3x^{2}-5x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\left(-13\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12\left(-13\right)}}{2\times 3}$$

Multiply -12 times -13.

$$x=\frac{-\left(-5\right)±\sqrt{25+156}}{2\times 3}$$

Add 25 to 156.

$$x=\frac{-\left(-5\right)±\sqrt{181}}{2\times 3}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{181}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{5±\sqrt{181}}{6}$$

Now solve the equation x=\frac{5±\sqrt{181}}{6} when ± is plus. Add 5 to \sqrt{181}.

$$x=\frac{\sqrt{181}+5}{6}$$

Now solve the equation x=\frac{5±\sqrt{181}}{6} when ± is minus. Subtract \sqrt{181} from 5.

$$x=\frac{5-\sqrt{181}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{181}+5}{6} x=\frac{5-\sqrt{181}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 5 by 2x+1.

$$10x+5+3\left(2x-6\right)=\left(-\left(x-7\right)\right)\times 3x$$

Use the distributive property to multiply 3 by 2x-6.

$$10x+5+6x-18=\left(-\left(x-7\right)\right)\times 3x$$

Combine 10x and 6x to get 16x.

$$16x+5-18=\left(-\left(x-7\right)\right)\times 3x$$

Subtract 18 from 5 to get -13.

$$16x-13=\left(-\left(x-7\right)\right)\times 3x$$

To find the opposite of x-7, find the opposite of each term.

$$16x-13=\left(-x-\left(-7\right)\right)\times 3x$$

The opposite of -7 is 7.

$$16x-13=\left(-x+7\right)\times 3x$$

Use the distributive property to multiply -x+7 by 3.

$$16x-13=\left(-3x+21\right)x$$

Use the distributive property to multiply -3x+21 by x.

$$16x-13=-3x^{2}+21x$$

Add 3x^{2} to both sides.

$$16x-13+3x^{2}=21x$$

Subtract 21x from both sides.

$$16x-13+3x^{2}-21x=0$$

Combine 16x and -21x to get -5x.

$$-5x-13+3x^{2}=0$$

Add 13 to both sides. Anything plus zero gives itself.

$$-5x+3x^{2}=13$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}-5x=13$$

Divide both sides by 3.

$$\frac{3x^{2}-5x}{3}=\frac{13}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{5}{3}x=\frac{13}{3}$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\frac{13}{3}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{13}{3}+\frac{25}{36}$$

Add \frac{13}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{181}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{181}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{181}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{181}}{6} x-\frac{5}{6}=-\frac{\sqrt{181}}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{\sqrt{181}+5}{6} x=\frac{5-\sqrt{181}}{6}$$