Solve 282-7x/1-50=3+7x/4+5x

$$28 \frac { 2 - 7 x } { 1 - 50 } = \frac { 3 + 7 x } { 4 + 5 x }$$

Solve for x

$$x=\frac{\sqrt{30209}-23}{280}\approx 0.538597731$$

$$x=\frac{-\sqrt{30209}-23}{280}\approx -0.702883445$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{4}{5} since division by zero is not defined. Multiply both sides of the equation by 5x+4.

$$28\left(-\frac{5}{49}x-\frac{4}{49}\right)\left(2-7x\right)=3+7x$$

Use the distributive property to multiply 28 by -\frac{5}{49}x-\frac{4}{49}.

$$\left(-\frac{20}{7}x-\frac{16}{7}\right)\left(2-7x\right)=3+7x$$

Use the distributive property to multiply -\frac{20}{7}x-\frac{16}{7} by 2-7x and combine like terms.

$$\frac{72}{7}x+20x^{2}-\frac{32}{7}=3+7x$$

Subtract 3 from both sides.

$$\frac{72}{7}x+20x^{2}-\frac{32}{7}-3=7x$$

Subtract 3 from -\frac{32}{7} to get -\frac{53}{7}.

$$\frac{72}{7}x+20x^{2}-\frac{53}{7}=7x$$

Subtract 7x from both sides.

$$\frac{72}{7}x+20x^{2}-\frac{53}{7}-7x=0$$

Combine \frac{72}{7}x and -7x to get \frac{23}{7}x.

$$\frac{23}{7}x+20x^{2}-\frac{53}{7}=0$$

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.

$$20x^{2}+\frac{23}{7}x-\frac{53}{7}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, \frac{23}{7} for b, and -\frac{53}{7} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{23}{7}±\sqrt{\left(\frac{23}{7}\right)^{2}-4\times 20\left(-\frac{53}{7}\right)}}{2\times 20}$$

Square \frac{23}{7} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{23}{7}±\sqrt{\frac{529}{49}-4\times 20\left(-\frac{53}{7}\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\frac{23}{7}±\sqrt{\frac{529}{49}-80\left(-\frac{53}{7}\right)}}{2\times 20}$$

Multiply -80 times -\frac{53}{7}.

$$x=\frac{-\frac{23}{7}±\sqrt{\frac{529}{49}+\frac{4240}{7}}}{2\times 20}$$

Add \frac{529}{49} to \frac{4240}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{23}{7}±\sqrt{\frac{30209}{49}}}{2\times 20}$$

Take the square root of \frac{30209}{49}.

$$x=\frac{-\frac{23}{7}±\frac{\sqrt{30209}}{7}}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-\frac{23}{7}±\frac{\sqrt{30209}}{7}}{40}$$

Now solve the equation x=\frac{-\frac{23}{7}±\frac{\sqrt{30209}}{7}}{40} when ± is plus. Add -\frac{23}{7} to \frac{\sqrt{30209}}{7}.

$$x=\frac{\sqrt{30209}-23}{7\times 40}$$

Divide \frac{-23+\sqrt{30209}}{7} by 40.

$$x=\frac{\sqrt{30209}-23}{280}$$

Now solve the equation x=\frac{-\frac{23}{7}±\frac{\sqrt{30209}}{7}}{40} when ± is minus. Subtract \frac{\sqrt{30209}}{7} from -\frac{23}{7}.

$$x=\frac{-\sqrt{30209}-23}{7\times 40}$$

Divide \frac{-23-\sqrt{30209}}{7} by 40.

$$x=\frac{-\sqrt{30209}-23}{280}$$

The equation is now solved.

$$x=\frac{\sqrt{30209}-23}{280} x=\frac{-\sqrt{30209}-23}{280}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{4}{5} since division by zero is not defined. Multiply both sides of the equation by 5x+4.

$$28\left(-\frac{5}{49}x-\frac{4}{49}\right)\left(2-7x\right)=3+7x$$

Use the distributive property to multiply 28 by -\frac{5}{49}x-\frac{4}{49}.

$$\left(-\frac{20}{7}x-\frac{16}{7}\right)\left(2-7x\right)=3+7x$$

Use the distributive property to multiply -\frac{20}{7}x-\frac{16}{7} by 2-7x and combine like terms.

$$\frac{72}{7}x+20x^{2}-\frac{32}{7}=3+7x$$

Subtract 7x from both sides.

$$\frac{72}{7}x+20x^{2}-\frac{32}{7}-7x=3$$

Combine \frac{72}{7}x and -7x to get \frac{23}{7}x.

$$\frac{23}{7}x+20x^{2}-\frac{32}{7}=3$$

Add \frac{32}{7} to both sides.

$$\frac{23}{7}x+20x^{2}=3+\frac{32}{7}$$

Add 3 and \frac{32}{7} to get \frac{53}{7}.

$$\frac{23}{7}x+20x^{2}=\frac{53}{7}$$

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.

$$20x^{2}+\frac{23}{7}x=\frac{53}{7}$$

Divide both sides by 20.

$$\frac{20x^{2}+\frac{23}{7}x}{20}=\frac{\frac{53}{7}}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{\frac{23}{7}}{20}x=\frac{\frac{53}{7}}{20}$$

Divide \frac{23}{7} by 20.

$$x^{2}+\frac{23}{140}x=\frac{\frac{53}{7}}{20}$$

Divide \frac{53}{7} by 20.

$$x^{2}+\frac{23}{140}x=\frac{53}{140}$$

Divide \frac{23}{140}, the coefficient of the x term, by 2 to get \frac{23}{280}. Then add the square of \frac{23}{280} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{23}{140}x+\left(\frac{23}{280}\right)^{2}=\frac{53}{140}+\left(\frac{23}{280}\right)^{2}$$

Square \frac{23}{280} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{23}{140}x+\frac{529}{78400}=\frac{53}{140}+\frac{529}{78400}$$

Add \frac{53}{140} to \frac{529}{78400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{23}{140}x+\frac{529}{78400}=\frac{30209}{78400}$$

Factor x^{2}+\frac{23}{140}x+\frac{529}{78400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{23}{280}\right)^{2}=\frac{30209}{78400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{23}{280}\right)^{2}}=\sqrt{\frac{30209}{78400}}$$

Simplify.

$$x+\frac{23}{280}=\frac{\sqrt{30209}}{280} x+\frac{23}{280}=-\frac{\sqrt{30209}}{280}$$

Subtract \frac{23}{280} from both sides of the equation.

$$x=\frac{\sqrt{30209}-23}{280} x=\frac{-\sqrt{30209}-23}{280}$$