Solve f(x)=3x^2-18x+10

$$f ( x ) = 3 x ^ { 2 } - 18 x + 10$$

Factor

$$3\left(x-\left(-\frac{\sqrt{51}}{3}+3\right)\right)\left(x-\left(\frac{\sqrt{51}}{3}+3\right)\right)$$

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.

$$3x^{2}-18x+10=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 10}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-\left(-18\right)±\sqrt{324-120}}{2\times 3}$$

Add 324 to -120.

$$x=\frac{-\left(-18\right)±\sqrt{204}}{2\times 3}$$

Take the square root of 204.

$$x=\frac{-\left(-18\right)±2\sqrt{51}}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{51}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±2\sqrt{51}}{6}$$

Now solve the equation x=\frac{18±2\sqrt{51}}{6} when ± is plus. Add 18 to 2\sqrt{51}.

$$x=\frac{2\sqrt{51}+18}{6}$$

Divide 18+2\sqrt{51} by 6.

$$x=\frac{\sqrt{51}}{3}+3$$

Now solve the equation x=\frac{18±2\sqrt{51}}{6} when ± is minus. Subtract 2\sqrt{51} from 18.

$$x=\frac{18-2\sqrt{51}}{6}$$

Divide 18-2\sqrt{51} by 6.

$$x=-\frac{\sqrt{51}}{3}+3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3+\frac{\sqrt{51}}{3} for x_{1} and 3-\frac{\sqrt{51}}{3} for x_{2}.

$$3x^{2}-18x+10=3\left(x-\left(\frac{\sqrt{51}}{3}+3\right)\right)\left(x-\left(-\frac{\sqrt{51}}{3}+3\right)\right)$$

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.This is achieved by dividing both sides of the equation by 3

$$x ^ 2 -6x +\frac{10}{3} = 0$$

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

$$r + s = 6 rs = \frac{10}{3}$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{3}

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

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

$$9 - u^2 = \frac{10}{3}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{10}{3}-9 = -\frac{17}{3}$$

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

$$u^2 = \frac{17}{3} u = \pm\sqrt{\frac{17}{3}} = \pm \frac{\sqrt{17}}{\sqrt{3}}$$

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

$$r =3 - \frac{\sqrt{17}}{\sqrt{3}} = 0.620 s = 3 + \frac{\sqrt{17}}{\sqrt{3}} = 5.380$$

Evaluate

$$3x^{2}-18x+10$$