Solve x:2(5x+9)=78

$$x : 2 ( 5 x + 9 ) = 78$$

Solve for x

$$x = \frac{\sqrt{3201} - 9}{10} \approx 4.757738064$$

$$x=\frac{-\sqrt{3201}-9}{10}\approx -6.557738064$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$x\left(5x+9\right)=156$$

Use the distributive property to multiply x by 5x+9.

$$5x^{2}+9x=156$$

Subtract 156 from both sides.

$$5x^{2}+9x-156=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 9 for b, and -156 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 5\left(-156\right)}}{2\times 5}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 5\left(-156\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-9±\sqrt{81-20\left(-156\right)}}{2\times 5}$$

Multiply -20 times -156.

$$x=\frac{-9±\sqrt{81+3120}}{2\times 5}$$

Add 81 to 3120.

$$x=\frac{-9±\sqrt{3201}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-9±\sqrt{3201}}{10}$$

Now solve the equation x=\frac{-9±\sqrt{3201}}{10} when ± is plus. Add -9 to \sqrt{3201}.

$$x=\frac{\sqrt{3201}-9}{10}$$

Now solve the equation x=\frac{-9±\sqrt{3201}}{10} when ± is minus. Subtract \sqrt{3201} from -9.

$$x=\frac{-\sqrt{3201}-9}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{3201}-9}{10} x=\frac{-\sqrt{3201}-9}{10}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$x\left(5x+9\right)=156$$

Use the distributive property to multiply x by 5x+9.

$$5x^{2}+9x=156$$

Divide both sides by 5.

$$\frac{5x^{2}+9x}{5}=\frac{156}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{9}{5}x=\frac{156}{5}$$

Divide \frac{9}{5}, the coefficient of the x term, by 2 to get \frac{9}{10}. Then add the square of \frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{5}x+\left(\frac{9}{10}\right)^{2}=\frac{156}{5}+\left(\frac{9}{10}\right)^{2}$$

Square \frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{5}x+\frac{81}{100}=\frac{156}{5}+\frac{81}{100}$$

Add \frac{156}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{5}x+\frac{81}{100}=\frac{3201}{100}$$

Factor x^{2}+\frac{9}{5}x+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{10}\right)^{2}=\frac{3201}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{10}\right)^{2}}=\sqrt{\frac{3201}{100}}$$

Simplify.

$$x+\frac{9}{10}=\frac{\sqrt{3201}}{10} x+\frac{9}{10}=-\frac{\sqrt{3201}}{10}$$

Subtract \frac{9}{10} from both sides of the equation.

$$x=\frac{\sqrt{3201}-9}{10} x=\frac{-\sqrt{3201}-9}{10}$$