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

$$\frac { x - 3 } { 5 } * \frac { x + 3 } { 4 } = \frac { 4 x + 1 } { 7 }$$

Solve for x

$$x = \frac{\sqrt{2181} + 40}{7} \approx 12.385882531$$

$$x=\frac{40-\sqrt{2181}}{7}\approx -0.957311102$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 140, the least common multiple of 5,4,7.

$$7\left(x-3\right)\left(x+3\right)=20\left(4x+1\right)$$

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

$$\left(7x-21\right)\left(x+3\right)=20\left(4x+1\right)$$

Apply the distributive property by multiplying each term of 7x-21 by each term of x+3.

$$7x^{2}+21x-21x-63=20\left(4x+1\right)$$

Combine 21x and -21x to get 0.

$$7x^{2}-63=20\left(4x+1\right)$$

Use the distributive property to multiply 20 by 4x+1.

$$7x^{2}-63=80x+20$$

Subtract 80x from both sides.

$$7x^{2}-63-80x=20$$

Subtract 20 from both sides.

$$7x^{2}-63-80x-20=0$$

Subtract 20 from -63 to get -83.

$$7x^{2}-83-80x=0$$

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.

$$7x^{2}-80x-83=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -80 for b, and -83 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 7\left(-83\right)}}{2\times 7}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 7\left(-83\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-80\right)±\sqrt{6400-28\left(-83\right)}}{2\times 7}$$

Multiply -28 times -83.

$$x=\frac{-\left(-80\right)±\sqrt{6400+2324}}{2\times 7}$$

Add 6400 to 2324.

$$x=\frac{-\left(-80\right)±\sqrt{8724}}{2\times 7}$$

Take the square root of 8724.

$$x=\frac{-\left(-80\right)±2\sqrt{2181}}{2\times 7}$$

The opposite of -80 is 80.

$$x=\frac{80±2\sqrt{2181}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{80±2\sqrt{2181}}{14}$$

Now solve the equation x=\frac{80±2\sqrt{2181}}{14} when ± is plus. Add 80 to 2\sqrt{2181}.

$$x=\frac{2\sqrt{2181}+80}{14}$$

Divide 80+2\sqrt{2181} by 14.

$$x=\frac{\sqrt{2181}+40}{7}$$

Now solve the equation x=\frac{80±2\sqrt{2181}}{14} when ± is minus. Subtract 2\sqrt{2181} from 80.

$$x=\frac{80-2\sqrt{2181}}{14}$$

Divide 80-2\sqrt{2181} by 14.

$$x=\frac{40-\sqrt{2181}}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{2181}+40}{7} x=\frac{40-\sqrt{2181}}{7}$$

Steps for Completing the Square

Multiply both sides of the equation by 140, the least common multiple of 5,4,7.

$$7\left(x-3\right)\left(x+3\right)=20\left(4x+1\right)$$

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

$$\left(7x-21\right)\left(x+3\right)=20\left(4x+1\right)$$

Apply the distributive property by multiplying each term of 7x-21 by each term of x+3.

$$7x^{2}+21x-21x-63=20\left(4x+1\right)$$

Combine 21x and -21x to get 0.

$$7x^{2}-63=20\left(4x+1\right)$$

Use the distributive property to multiply 20 by 4x+1.

$$7x^{2}-63=80x+20$$

Subtract 80x from both sides.

$$7x^{2}-63-80x=20$$

Add 63 to both sides.

$$7x^{2}-80x=20+63$$

Add 20 and 63 to get 83.

$$7x^{2}-80x=83$$

Divide both sides by 7.

$$\frac{7x^{2}-80x}{7}=\frac{83}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{80}{7}x=\frac{83}{7}$$

Divide -\frac{80}{7}, the coefficient of the x term, by 2 to get -\frac{40}{7}. Then add the square of -\frac{40}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{80}{7}x+\left(-\frac{40}{7}\right)^{2}=\frac{83}{7}+\left(-\frac{40}{7}\right)^{2}$$

Square -\frac{40}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{80}{7}x+\frac{1600}{49}=\frac{83}{7}+\frac{1600}{49}$$

Add \frac{83}{7} to \frac{1600}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{80}{7}x+\frac{1600}{49}=\frac{2181}{49}$$

Factor x^{2}-\frac{80}{7}x+\frac{1600}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{40}{7}\right)^{2}=\frac{2181}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{40}{7}\right)^{2}}=\sqrt{\frac{2181}{49}}$$

Simplify.

$$x-\frac{40}{7}=\frac{\sqrt{2181}}{7} x-\frac{40}{7}=-\frac{\sqrt{2181}}{7}$$

Add \frac{40}{7} to both sides of the equation.

$$x=\frac{\sqrt{2181}+40}{7} x=\frac{40-\sqrt{2181}}{7}$$