Solve -45x^2+27x+36

$$- 45 x ^ { 2 } + 27 x + 36$$

Factor

$$-45\left(x-\frac{3-\sqrt{89}}{10}\right)\left(x-\frac{\sqrt{89}+3}{10}\right)$$

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.

$$-45x^{2}+27x+36=0$$

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.

$$x=\frac{-27±\sqrt{27^{2}-4\left(-45\right)\times 36}}{2\left(-45\right)}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\left(-45\right)\times 36}}{2\left(-45\right)}$$

Multiply -4 times -45.

$$x=\frac{-27±\sqrt{729+180\times 36}}{2\left(-45\right)}$$

Multiply 180 times 36.

$$x=\frac{-27±\sqrt{729+6480}}{2\left(-45\right)}$$

Add 729 to 6480.

$$x=\frac{-27±\sqrt{7209}}{2\left(-45\right)}$$

Take the square root of 7209.

$$x=\frac{-27±9\sqrt{89}}{2\left(-45\right)}$$

Multiply 2 times -45.

$$x=\frac{-27±9\sqrt{89}}{-90}$$

Now solve the equation x=\frac{-27±9\sqrt{89}}{-90} when ± is plus. Add -27 to 9\sqrt{89}.

$$x=\frac{9\sqrt{89}-27}{-90}$$

Divide -27+9\sqrt{89} by -90.

$$x=\frac{3-\sqrt{89}}{10}$$

Now solve the equation x=\frac{-27±9\sqrt{89}}{-90} when ± is minus. Subtract 9\sqrt{89} from -27.

$$x=\frac{-9\sqrt{89}-27}{-90}$$

Divide -27-9\sqrt{89} by -90.

$$x=\frac{\sqrt{89}+3}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3-\sqrt{89}}{10} for x_{1} and \frac{3+\sqrt{89}}{10} for x_{2}.

$$-45x^{2}+27x+36=-45\left(x-\frac{3-\sqrt{89}}{10}\right)\left(x-\frac{\sqrt{89}+3}{10}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -\frac{3}{5}x -\frac{4}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{3}{5} rs = -\frac{4}{5}$$

Two numbers r and s sum up to \frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{5} = \frac{3}{10}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{3}{10} - u s = \frac{3}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}

$$(\frac{3}{10} - u) (\frac{3}{10} + u) = -\frac{4}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{100} - u^2 = -\frac{4}{5}$$

Simplify the expression by subtracting \frac{9}{100} on both sides

$$-u^2 = -\frac{4}{5}-\frac{9}{100} = -\frac{89}{100}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{89}{100} u = \pm\sqrt{\frac{89}{100}} = \pm \frac{\sqrt{89}}{10}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{3}{10} - \frac{\sqrt{89}}{10} = -0.643 s = \frac{3}{10} + \frac{\sqrt{89}}{10} = 1.243$$

Evaluate

$$36+27x-45x^{2}$$