Solve (3x+1+2x)(1-2x)=61x+50(x+3)

$$( 3 x + 1 + 2 x ) ( 1 - 2 x ) = 61 x + 50 ( x + 3 )$$

Solve for x

$$x=\frac{\sqrt{1426}}{10}-\frac{27}{5}\approx -1.623758482$$

$$x=-\frac{\sqrt{1426}}{10}-\frac{27}{5}\approx -9.176241518$$

Steps Using the Quadratic Formula

Combine 3x and 2x to get 5x.

$$\left(5x+1\right)\left(1-2x\right)=61x+50\left(x+3\right)$$

Use the distributive property to multiply 5x+1 by 1-2x and combine like terms.

$$3x-10x^{2}+1=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$3x-10x^{2}+1=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$3x-10x^{2}+1=111x+150$$

Subtract 111x from both sides.

$$3x-10x^{2}+1-111x=150$$

Combine 3x and -111x to get -108x.

$$-108x-10x^{2}+1=150$$

Subtract 150 from both sides.

$$-108x-10x^{2}+1-150=0$$

Subtract 150 from 1 to get -149.

$$-108x-10x^{2}-149=0$$

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.

$$-10x^{2}-108x-149=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -108 for b, and -149 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-108\right)±\sqrt{\left(-108\right)^{2}-4\left(-10\right)\left(-149\right)}}{2\left(-10\right)}$$

Square -108.

$$x=\frac{-\left(-108\right)±\sqrt{11664-4\left(-10\right)\left(-149\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-108\right)±\sqrt{11664+40\left(-149\right)}}{2\left(-10\right)}$$

Multiply 40 times -149.

$$x=\frac{-\left(-108\right)±\sqrt{11664-5960}}{2\left(-10\right)}$$

Add 11664 to -5960.

$$x=\frac{-\left(-108\right)±\sqrt{5704}}{2\left(-10\right)}$$

Take the square root of 5704.

$$x=\frac{-\left(-108\right)±2\sqrt{1426}}{2\left(-10\right)}$$

The opposite of -108 is 108.

$$x=\frac{108±2\sqrt{1426}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{108±2\sqrt{1426}}{-20}$$

Now solve the equation x=\frac{108±2\sqrt{1426}}{-20} when ± is plus. Add 108 to 2\sqrt{1426}.

$$x=\frac{2\sqrt{1426}+108}{-20}$$

Divide 108+2\sqrt{1426} by -20.

$$x=-\frac{\sqrt{1426}}{10}-\frac{27}{5}$$

Now solve the equation x=\frac{108±2\sqrt{1426}}{-20} when ± is minus. Subtract 2\sqrt{1426} from 108.

$$x=\frac{108-2\sqrt{1426}}{-20}$$

Divide 108-2\sqrt{1426} by -20.

$$x=\frac{\sqrt{1426}}{10}-\frac{27}{5}$$

The equation is now solved.

$$x=-\frac{\sqrt{1426}}{10}-\frac{27}{5} x=\frac{\sqrt{1426}}{10}-\frac{27}{5}$$

Steps for Completing the Square

Combine 3x and 2x to get 5x.

$$\left(5x+1\right)\left(1-2x\right)=61x+50\left(x+3\right)$$

Use the distributive property to multiply 5x+1 by 1-2x and combine like terms.

$$3x-10x^{2}+1=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$3x-10x^{2}+1=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$3x-10x^{2}+1=111x+150$$

Subtract 111x from both sides.

$$3x-10x^{2}+1-111x=150$$

Combine 3x and -111x to get -108x.

$$-108x-10x^{2}+1=150$$

Subtract 1 from both sides.

$$-108x-10x^{2}=150-1$$

Subtract 1 from 150 to get 149.

$$-108x-10x^{2}=149$$

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.

$$-10x^{2}-108x=149$$

Divide both sides by -10.

$$\frac{-10x^{2}-108x}{-10}=\frac{149}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\left(-\frac{108}{-10}\right)x=\frac{149}{-10}$$

Reduce the fraction \frac{-108}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{54}{5}x=\frac{149}{-10}$$

Divide 149 by -10.

$$x^{2}+\frac{54}{5}x=-\frac{149}{10}$$

Divide \frac{54}{5}, the coefficient of the x term, by 2 to get \frac{27}{5}. Then add the square of \frac{27}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{54}{5}x+\left(\frac{27}{5}\right)^{2}=-\frac{149}{10}+\left(\frac{27}{5}\right)^{2}$$

Square \frac{27}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{54}{5}x+\frac{729}{25}=-\frac{149}{10}+\frac{729}{25}$$

Add -\frac{149}{10} to \frac{729}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{54}{5}x+\frac{729}{25}=\frac{713}{50}$$

Factor x^{2}+\frac{54}{5}x+\frac{729}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{27}{5}\right)^{2}=\frac{713}{50}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{5}\right)^{2}}=\sqrt{\frac{713}{50}}$$

Simplify.

$$x+\frac{27}{5}=\frac{\sqrt{1426}}{10} x+\frac{27}{5}=-\frac{\sqrt{1426}}{10}$$

Subtract \frac{27}{5} from both sides of the equation.

$$x=\frac{\sqrt{1426}}{10}-\frac{27}{5} x=-\frac{\sqrt{1426}}{10}-\frac{27}{5}$$