Solve 2(x-5)1/8(9x+5)=10

$$2(x-5) \frac{ 1 }{ 8 } (9x+5)=10$$

Solve for x

$$x = \frac{\sqrt{985} + 20}{9} \approx 5.709412184$$

$$x=\frac{20-\sqrt{985}}{9}\approx -1.264967739$$

Steps Using the Quadratic Formula

Multiply 2 and \frac{1}{8} to get \frac{1}{4}.

$$\frac{1}{4}\left(x-5\right)\left(9x+5\right)=10$$

Use the distributive property to multiply \frac{1}{4} by x-5.

$$\left(\frac{1}{4}x-\frac{5}{4}\right)\left(9x+5\right)=10$$

Use the distributive property to multiply \frac{1}{4}x-\frac{5}{4} by 9x+5 and combine like terms.

$$\frac{9}{4}x^{2}-10x-\frac{25}{4}=10$$

Subtract 10 from both sides.

$$\frac{9}{4}x^{2}-10x-\frac{25}{4}-10=0$$

Subtract 10 from -\frac{25}{4} to get -\frac{65}{4}.

$$\frac{9}{4}x^{2}-10x-\frac{65}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{9}{4} for a, -10 for b, and -\frac{65}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times \frac{9}{4}\left(-\frac{65}{4}\right)}}{2\times \frac{9}{4}}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times \frac{9}{4}\left(-\frac{65}{4}\right)}}{2\times \frac{9}{4}}$$

Multiply -4 times \frac{9}{4}.

$$x=\frac{-\left(-10\right)±\sqrt{100-9\left(-\frac{65}{4}\right)}}{2\times \frac{9}{4}}$$

Multiply -9 times -\frac{65}{4}.

$$x=\frac{-\left(-10\right)±\sqrt{100+\frac{585}{4}}}{2\times \frac{9}{4}}$$

Add 100 to \frac{585}{4}.

$$x=\frac{-\left(-10\right)±\sqrt{\frac{985}{4}}}{2\times \frac{9}{4}}$$

Take the square root of \frac{985}{4}.

$$x=\frac{-\left(-10\right)±\frac{\sqrt{985}}{2}}{2\times \frac{9}{4}}$$

The opposite of -10 is 10.

$$x=\frac{10±\frac{\sqrt{985}}{2}}{2\times \frac{9}{4}}$$

Multiply 2 times \frac{9}{4}.

$$x=\frac{10±\frac{\sqrt{985}}{2}}{\frac{9}{2}}$$

Now solve the equation x=\frac{10±\frac{\sqrt{985}}{2}}{\frac{9}{2}} when ± is plus. Add 10 to \frac{\sqrt{985}}{2}.

$$x=\frac{\frac{\sqrt{985}}{2}+10}{\frac{9}{2}}$$

Divide 10+\frac{\sqrt{985}}{2} by \frac{9}{2} by multiplying 10+\frac{\sqrt{985}}{2} by the reciprocal of \frac{9}{2}.

$$x=\frac{\sqrt{985}+20}{9}$$

Now solve the equation x=\frac{10±\frac{\sqrt{985}}{2}}{\frac{9}{2}} when ± is minus. Subtract \frac{\sqrt{985}}{2} from 10.

$$x=\frac{-\frac{\sqrt{985}}{2}+10}{\frac{9}{2}}$$

Divide 10-\frac{\sqrt{985}}{2} by \frac{9}{2} by multiplying 10-\frac{\sqrt{985}}{2} by the reciprocal of \frac{9}{2}.

$$x=\frac{20-\sqrt{985}}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{985}+20}{9} x=\frac{20-\sqrt{985}}{9}$$

Steps for Completing the Square

Multiply 2 and \frac{1}{8} to get \frac{1}{4}.

$$\frac{1}{4}\left(x-5\right)\left(9x+5\right)=10$$

Use the distributive property to multiply \frac{1}{4} by x-5.

$$\left(\frac{1}{4}x-\frac{5}{4}\right)\left(9x+5\right)=10$$

Use the distributive property to multiply \frac{1}{4}x-\frac{5}{4} by 9x+5 and combine like terms.

$$\frac{9}{4}x^{2}-10x-\frac{25}{4}=10$$

Add \frac{25}{4} to both sides.

$$\frac{9}{4}x^{2}-10x=10+\frac{25}{4}$$

Add 10 and \frac{25}{4} to get \frac{65}{4}.

$$\frac{9}{4}x^{2}-10x=\frac{65}{4}$$

Divide both sides of the equation by \frac{9}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{4}x^{2}-10x}{\frac{9}{4}}=\frac{\frac{65}{4}}{\frac{9}{4}}$$

Dividing by \frac{9}{4} undoes the multiplication by \frac{9}{4}.

$$x^{2}+\left(-\frac{10}{\frac{9}{4}}\right)x=\frac{\frac{65}{4}}{\frac{9}{4}}$$

Divide -10 by \frac{9}{4} by multiplying -10 by the reciprocal of \frac{9}{4}.

$$x^{2}-\frac{40}{9}x=\frac{\frac{65}{4}}{\frac{9}{4}}$$

Divide \frac{65}{4} by \frac{9}{4} by multiplying \frac{65}{4} by the reciprocal of \frac{9}{4}.

$$x^{2}-\frac{40}{9}x=\frac{65}{9}$$

Divide -\frac{40}{9}, the coefficient of the x term, by 2 to get -\frac{20}{9}. Then add the square of -\frac{20}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{40}{9}x+\left(-\frac{20}{9}\right)^{2}=\frac{65}{9}+\left(-\frac{20}{9}\right)^{2}$$

Square -\frac{20}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{40}{9}x+\frac{400}{81}=\frac{65}{9}+\frac{400}{81}$$

Add \frac{65}{9} to \frac{400}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{40}{9}x+\frac{400}{81}=\frac{985}{81}$$

Factor x^{2}-\frac{40}{9}x+\frac{400}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{20}{9}\right)^{2}=\frac{985}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{20}{9}\right)^{2}}=\sqrt{\frac{985}{81}}$$

Simplify.

$$x-\frac{20}{9}=\frac{\sqrt{985}}{9} x-\frac{20}{9}=-\frac{\sqrt{985}}{9}$$

Add \frac{20}{9} to both sides of the equation.

$$x=\frac{\sqrt{985}+20}{9} x=\frac{20-\sqrt{985}}{9}$$