Solve 4x+1/3*2x-1/2-1x-7/5=6

$$\frac { 4 x + 1 } { 3 } \cdot \frac { 2 x - 1 } { 2 } - \frac { 1 x - 7 } { 5 } = 6$$

Solve for x

$$x=\frac{\sqrt{1446}}{20}+\frac{1}{5}\approx 2.101315334$$

$$x=-\frac{\sqrt{1446}}{20}+\frac{1}{5}\approx -1.701315334$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 30, the least common multiple of 3,2,5.

$$5\left(4x+1\right)\left(2x-1\right)-6\left(x-7\right)=180$$

Use the distributive property to multiply 5 by 4x+1.

$$\left(20x+5\right)\left(2x-1\right)-6\left(x-7\right)=180$$

Apply the distributive property by multiplying each term of 20x+5 by each term of 2x-1.

$$40x^{2}-20x+10x-5-6\left(x-7\right)=180$$

Combine -20x and 10x to get -10x.

$$40x^{2}-10x-5-6\left(x-7\right)=180$$

Use the distributive property to multiply -6 by 1x-7.

$$40x^{2}-10x-5-6x+42=180$$

Combine -10x and -6x to get -16x.

$$40x^{2}-16x-5+42=180$$

Add -5 and 42 to get 37.

$$40x^{2}-16x+37=180$$

Subtract 180 from both sides.

$$40x^{2}-16x+37-180=0$$

Subtract 180 from 37 to get -143.

$$40x^{2}-16x-143=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -16 for b, and -143 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 40\left(-143\right)}}{2\times 40}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 40\left(-143\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-16\right)±\sqrt{256-160\left(-143\right)}}{2\times 40}$$

Multiply -160 times -143.

$$x=\frac{-\left(-16\right)±\sqrt{256+22880}}{2\times 40}$$

Add 256 to 22880.

$$x=\frac{-\left(-16\right)±\sqrt{23136}}{2\times 40}$$

Take the square root of 23136.

$$x=\frac{-\left(-16\right)±4\sqrt{1446}}{2\times 40}$$

The opposite of -16 is 16.

$$x=\frac{16±4\sqrt{1446}}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{16±4\sqrt{1446}}{80}$$

Now solve the equation x=\frac{16±4\sqrt{1446}}{80} when ± is plus. Add 16 to 4\sqrt{1446}.

$$x=\frac{4\sqrt{1446}+16}{80}$$

Divide 16+4\sqrt{1446} by 80.

$$x=\frac{\sqrt{1446}}{20}+\frac{1}{5}$$

Now solve the equation x=\frac{16±4\sqrt{1446}}{80} when ± is minus. Subtract 4\sqrt{1446} from 16.

$$x=\frac{16-4\sqrt{1446}}{80}$$

Divide 16-4\sqrt{1446} by 80.

$$x=-\frac{\sqrt{1446}}{20}+\frac{1}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{1446}}{20}+\frac{1}{5} x=-\frac{\sqrt{1446}}{20}+\frac{1}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 30, the least common multiple of 3,2,5.

$$5\left(4x+1\right)\left(2x-1\right)-6\left(x-7\right)=180$$

Use the distributive property to multiply 5 by 4x+1.

$$\left(20x+5\right)\left(2x-1\right)-6\left(x-7\right)=180$$

Apply the distributive property by multiplying each term of 20x+5 by each term of 2x-1.

$$40x^{2}-20x+10x-5-6\left(x-7\right)=180$$

Combine -20x and 10x to get -10x.

$$40x^{2}-10x-5-6\left(x-7\right)=180$$

Use the distributive property to multiply -6 by 1x-7.

$$40x^{2}-10x-5-6x+42=180$$

Combine -10x and -6x to get -16x.

$$40x^{2}-16x-5+42=180$$

Add -5 and 42 to get 37.

$$40x^{2}-16x+37=180$$

Subtract 37 from both sides.

$$40x^{2}-16x=180-37$$

Subtract 37 from 180 to get 143.

$$40x^{2}-16x=143$$

Divide both sides by 40.

$$\frac{40x^{2}-16x}{40}=\frac{143}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{16}{40}\right)x=\frac{143}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{2}{5}x=\frac{143}{40}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{143}{40}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{143}{40}+\frac{1}{25}$$

Add \frac{143}{40} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{723}{200}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{723}{200}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{723}{200}}$$

Simplify.

$$x-\frac{1}{5}=\frac{\sqrt{1446}}{20} x-\frac{1}{5}=-\frac{\sqrt{1446}}{20}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{\sqrt{1446}}{20}+\frac{1}{5} x=-\frac{\sqrt{1446}}{20}+\frac{1}{5}$$