Solve 7a*5/4a=10a

$$7 a \cdot \frac { 5 } { 4 } a = 10 a$$

Solve for a

$$a = \frac{8}{7} = 1\frac{1}{7} \approx 1.142857143$$

$$a=0$$

Steps Using Factoring

Multiply a and a to get a^{2}.

$$7a^{2}\times \frac{5}{4}=10a$$

Express 7\times \frac{5}{4} as a single fraction.

$$\frac{7\times 5}{4}a^{2}=10a$$

Multiply 7 and 5 to get 35.

$$\frac{35}{4}a^{2}=10a$$

Subtract 10a from both sides.

$$\frac{35}{4}a^{2}-10a=0$$

Factor out a.

$$a\left(\frac{35}{4}a-10\right)=0$$

To find equation solutions, solve a=0 and \frac{35a}{4}-10=0.

$$a=0 a=\frac{8}{7}$$

Steps Using the Quadratic Formula

Multiply a and a to get a^{2}.

$$7a^{2}\times \frac{5}{4}=10a$$

Express 7\times \frac{5}{4} as a single fraction.

$$\frac{7\times 5}{4}a^{2}=10a$$

Multiply 7 and 5 to get 35.

$$\frac{35}{4}a^{2}=10a$$

Subtract 10a from both sides.

$$\frac{35}{4}a^{2}-10a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{35}{4} for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times \frac{35}{4}}$$

Take the square root of \left(-10\right)^{2}.

$$a=\frac{-\left(-10\right)±10}{2\times \frac{35}{4}}$$

The opposite of -10 is 10.

$$a=\frac{10±10}{2\times \frac{35}{4}}$$

Multiply 2 times \frac{35}{4}.

$$a=\frac{10±10}{\frac{35}{2}}$$

Now solve the equation a=\frac{10±10}{\frac{35}{2}} when ± is plus. Add 10 to 10.

$$a=\frac{20}{\frac{35}{2}}$$

Divide 20 by \frac{35}{2} by multiplying 20 by the reciprocal of \frac{35}{2}.

$$a=\frac{8}{7}$$

Now solve the equation a=\frac{10±10}{\frac{35}{2}} when ± is minus. Subtract 10 from 10.

$$a=\frac{0}{\frac{35}{2}}$$

Divide 0 by \frac{35}{2} by multiplying 0 by the reciprocal of \frac{35}{2}.

$$a=0$$

The equation is now solved.

$$a=\frac{8}{7} a=0$$

Steps for Completing the Square

Multiply a and a to get a^{2}.

$$7a^{2}\times \frac{5}{4}=10a$$

Express 7\times \frac{5}{4} as a single fraction.

$$\frac{7\times 5}{4}a^{2}=10a$$

Multiply 7 and 5 to get 35.

$$\frac{35}{4}a^{2}=10a$$

Subtract 10a from both sides.

$$\frac{35}{4}a^{2}-10a=0$$

Divide both sides of the equation by \frac{35}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{35}{4}a^{2}-10a}{\frac{35}{4}}=\frac{0}{\frac{35}{4}}$$

Dividing by \frac{35}{4} undoes the multiplication by \frac{35}{4}.

$$a^{2}+\left(-\frac{10}{\frac{35}{4}}\right)a=\frac{0}{\frac{35}{4}}$$

Divide -10 by \frac{35}{4} by multiplying -10 by the reciprocal of \frac{35}{4}.

$$a^{2}-\frac{8}{7}a=\frac{0}{\frac{35}{4}}$$

Divide 0 by \frac{35}{4} by multiplying 0 by the reciprocal of \frac{35}{4}.

$$a^{2}-\frac{8}{7}a=0$$

Divide -\frac{8}{7}, the coefficient of the x term, by 2 to get -\frac{4}{7}. Then add the square of -\frac{4}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{8}{7}a+\left(-\frac{4}{7}\right)^{2}=\left(-\frac{4}{7}\right)^{2}$$

Square -\frac{4}{7} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{8}{7}a+\frac{16}{49}=\frac{16}{49}$$

Factor a^{2}-\frac{8}{7}a+\frac{16}{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}.

$$\left(a-\frac{4}{7}\right)^{2}=\frac{16}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{4}{7}\right)^{2}}=\sqrt{\frac{16}{49}}$$

Simplify.

$$a-\frac{4}{7}=\frac{4}{7} a-\frac{4}{7}=-\frac{4}{7}$$

Add \frac{4}{7} to both sides of the equation.

$$a=\frac{8}{7} a=0$$