Solve -3x^2+15x+2

$$- 3 x ^ { 2 } + 15 x + 2$$

Factor

$$-3\left(x-\left(-\frac{\sqrt{249}}{6}+\frac{5}{2}\right)\right)\left(x-\left(\frac{\sqrt{249}}{6}+\frac{5}{2}\right)\right)$$

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.

$$-3x^{2}+15x+2=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-15±\sqrt{225+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-15±\sqrt{225+24}}{2\left(-3\right)}$$

Add 225 to 24.

$$x=\frac{-15±\sqrt{249}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-15±\sqrt{249}}{-6}$$

Now solve the equation x=\frac{-15±\sqrt{249}}{-6} when ± is plus. Add -15 to \sqrt{249}.

$$x=\frac{\sqrt{249}-15}{-6}$$

Divide -15+\sqrt{249} by -6.

$$x=-\frac{\sqrt{249}}{6}+\frac{5}{2}$$

Now solve the equation x=\frac{-15±\sqrt{249}}{-6} when ± is minus. Subtract \sqrt{249} from -15.

$$x=\frac{-\sqrt{249}-15}{-6}$$

Divide -15-\sqrt{249} by -6.

$$x=\frac{\sqrt{249}}{6}+\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2}-\frac{\sqrt{249}}{6} for x_{1} and \frac{5}{2}+\frac{\sqrt{249}}{6} for x_{2}.

$$-3x^{2}+15x+2=-3\left(x-\left(-\frac{\sqrt{249}}{6}+\frac{5}{2}\right)\right)\left(x-\left(\frac{\sqrt{249}}{6}+\frac{5}{2}\right)\right)$$

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.

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

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

$$r + s = 5 rs = -\frac{2}{3}$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{2}. 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>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -\frac{2}{3}$$

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

$$\frac{25}{4} - u^2 = -\frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -\frac{2}{3}-\frac{25}{4} = -\frac{83}{12}$$

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

$$u^2 = \frac{83}{12} u = \pm\sqrt{\frac{83}{12}} = \pm \frac{\sqrt{83}}{\sqrt{12}}$$

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

$$r =\frac{5}{2} - \frac{\sqrt{83}}{\sqrt{12}} = -0.130 s = \frac{5}{2} + \frac{\sqrt{83}}{\sqrt{12}} = 5.130$$

Evaluate

$$2+15x-3x^{2}$$