Solve -4n^2+15n+9

$$- 4 n ^ { 2 } + 15 n + 9$$

Factor

$$-4\left(n-\frac{15-3\sqrt{41}}{8}\right)\left(n-\frac{3\sqrt{41}+15}{8}\right)$$

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.

$$-4n^{2}+15n+9=0$$

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.

$$n=\frac{-15±\sqrt{15^{2}-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Square 15.

$$n=\frac{-15±\sqrt{225-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$n=\frac{-15±\sqrt{225+16\times 9}}{2\left(-4\right)}$$

Multiply 16 times 9.

$$n=\frac{-15±\sqrt{225+144}}{2\left(-4\right)}$$

Add 225 to 144.

$$n=\frac{-15±\sqrt{369}}{2\left(-4\right)}$$

Take the square root of 369.

$$n=\frac{-15±3\sqrt{41}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$n=\frac{-15±3\sqrt{41}}{-8}$$

Now solve the equation n=\frac{-15±3\sqrt{41}}{-8} when ± is plus. Add -15 to 3\sqrt{41}.

$$n=\frac{3\sqrt{41}-15}{-8}$$

Divide -15+3\sqrt{41} by -8.

$$n=\frac{15-3\sqrt{41}}{8}$$

Now solve the equation n=\frac{-15±3\sqrt{41}}{-8} when ± is minus. Subtract 3\sqrt{41} from -15.

$$n=\frac{-3\sqrt{41}-15}{-8}$$

Divide -15-3\sqrt{41} by -8.

$$n=\frac{3\sqrt{41}+15}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{15-3\sqrt{41}}{8} for x_{1} and \frac{15+3\sqrt{41}}{8} for x_{2}.

$$-4n^{2}+15n+9=-4\left(n-\frac{15-3\sqrt{41}}{8}\right)\left(n-\frac{3\sqrt{41}+15}{8}\right)$$

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.

$$x ^ 2 -\frac{15}{4}x -\frac{9}{4} = 0$$

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

$$r + s = \frac{15}{4} rs = -\frac{9}{4}$$

Two numbers r and s sum up to \frac{15}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{15}{4} = \frac{15}{8}. 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>

$$r = \frac{15}{8} - u s = \frac{15}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{4}

$$(\frac{15}{8} - u) (\frac{15}{8} + u) = -\frac{9}{4}$$

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

$$\frac{225}{64} - u^2 = -\frac{9}{4}$$

Simplify the expression by subtracting \frac{225}{64} on both sides

$$-u^2 = -\frac{9}{4}-\frac{225}{64} = -\frac{369}{64}$$

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

$$u^2 = \frac{369}{64} u = \pm\sqrt{\frac{369}{64}} = \pm \frac{\sqrt{369}}{8}$$

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

$$r =\frac{15}{8} - \frac{\sqrt{369}}{8} = -0.526 s = \frac{15}{8} + \frac{\sqrt{369}}{8} = 4.276$$

Evaluate

$$9+15n-4n^{2}$$