Solve left(9x+5right)left(-9x+50right)=736

$$\left( 9x+5 \right) \left( -9x+50 \right) = 736$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9x+5 by -9x+50 and combine like terms.

$$-81x^{2}+405x+250=736$$

Subtract 736 from both sides.

$$-81x^{2}+405x+250-736=0$$

Subtract 736 from 250 to get -486.

$$-81x^{2}+405x-486=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -81 for a, 405 for b, and -486 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-405±\sqrt{405^{2}-4\left(-81\right)\left(-486\right)}}{2\left(-81\right)}$$

Square 405.

$$x=\frac{-405±\sqrt{164025-4\left(-81\right)\left(-486\right)}}{2\left(-81\right)}$$

Multiply -4 times -81.

$$x=\frac{-405±\sqrt{164025+324\left(-486\right)}}{2\left(-81\right)}$$

Multiply 324 times -486.

$$x=\frac{-405±\sqrt{164025-157464}}{2\left(-81\right)}$$

Add 164025 to -157464.

$$x=\frac{-405±\sqrt{6561}}{2\left(-81\right)}$$

Take the square root of 6561.

$$x=\frac{-405±81}{2\left(-81\right)}$$

Multiply 2 times -81.

$$x=\frac{-405±81}{-162}$$

Now solve the equation x=\frac{-405±81}{-162} when ± is plus. Add -405 to 81.

$$x=-\frac{324}{-162}$$

Divide -324 by -162.

$$x=2$$

Now solve the equation x=\frac{-405±81}{-162} when ± is minus. Subtract 81 from -405.

$$x=-\frac{486}{-162}$$

Divide -486 by -162.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Steps for Completing the Square

Use the distributive property to multiply 9x+5 by -9x+50 and combine like terms.

$$-81x^{2}+405x+250=736$$

Subtract 250 from both sides.

$$-81x^{2}+405x=736-250$$

Subtract 250 from 736 to get 486.

$$-81x^{2}+405x=486$$

Divide both sides by -81.

$$\frac{-81x^{2}+405x}{-81}=\frac{486}{-81}$$

Dividing by -81 undoes the multiplication by -81.

$$x^{2}+\frac{405}{-81}x=\frac{486}{-81}$$

Divide 405 by -81.

$$x^{2}-5x=\frac{486}{-81}$$

Divide 486 by -81.

$$x^{2}-5x=-6$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$