Solve 81x-45x^2=

$$81 x - 45 x ^ { 2 } =$$

Factor

$$9x\left(9-5x\right)$$

Solution Steps

Factor out 9.

$$9\left(9x-5x^{2}\right)$$

Consider 9x-5x^{2}. Factor out x.

$$x\left(9-5x\right)$$

Rewrite the complete factored expression.

$$9x\left(-5x+9\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-45x^{2}+81x=0$$

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.

$$x=\frac{-81±\sqrt{81^{2}}}{2\left(-45\right)}$$

Take the square root of 81^{2}.

$$x=\frac{-81±81}{2\left(-45\right)}$$

Multiply 2 times -45.

$$x=\frac{-81±81}{-90}$$

Now solve the equation x=\frac{-81±81}{-90} when ± is plus. Add -81 to 81.

$$x=\frac{0}{-90}$$

Divide 0 by -90.

$$x=0$$

Now solve the equation x=\frac{-81±81}{-90} when ± is minus. Subtract 81 from -81.

$$x=-\frac{162}{-90}$$

Reduce the fraction \frac{-162}{-90} to lowest terms by extracting and canceling out 18.

$$x=\frac{9}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and \frac{9}{5} for x_{2}.

$$-45x^{2}+81x=-45x\left(x-\frac{9}{5}\right)$$

Subtract \frac{9}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-45x^{2}+81x=-45x\times \frac{-5x+9}{-5}$$

Cancel out 5, the greatest common factor in -45 and -5.

$$-45x^{2}+81x=9x\left(-5x+9\right)$$

Evaluate

$$9x\left(9-5x\right)$$