Solve f(t)=-3t^2+12t-1

$$f ( t ) = - 3 t ^ { 2 } + 12 t - 1$$

Factor

$$-3\left(t-\left(-\frac{\sqrt{33}}{3}+2\right)\right)\left(t-\left(\frac{\sqrt{33}}{3}+2\right)\right)$$

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.

$$-3t^{2}+12t-1=0$$

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.

$$t=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Square 12.

$$t=\frac{-12±\sqrt{144-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$t=\frac{-12±\sqrt{144+12\left(-1\right)}}{2\left(-3\right)}$$

Multiply 12 times -1.

$$t=\frac{-12±\sqrt{144-12}}{2\left(-3\right)}$$

Add 144 to -12.

$$t=\frac{-12±\sqrt{132}}{2\left(-3\right)}$$

Take the square root of 132.

$$t=\frac{-12±2\sqrt{33}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$t=\frac{-12±2\sqrt{33}}{-6}$$

Now solve the equation t=\frac{-12±2\sqrt{33}}{-6} when ± is plus. Add -12 to 2\sqrt{33}.

$$t=\frac{2\sqrt{33}-12}{-6}$$

Divide -12+2\sqrt{33} by -6.

$$t=-\frac{\sqrt{33}}{3}+2$$

Now solve the equation t=\frac{-12±2\sqrt{33}}{-6} when ± is minus. Subtract 2\sqrt{33} from -12.

$$t=\frac{-2\sqrt{33}-12}{-6}$$

Divide -12-2\sqrt{33} by -6.

$$t=\frac{\sqrt{33}}{3}+2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2-\frac{\sqrt{33}}{3} for x_{1} and 2+\frac{\sqrt{33}}{3} for x_{2}.

$$-3t^{2}+12t-1=-3\left(t-\left(-\frac{\sqrt{33}}{3}+2\right)\right)\left(t-\left(\frac{\sqrt{33}}{3}+2\right)\right)$$

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.

$$x ^ 2 -4x +\frac{1}{3} = 0$$

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

$$r + s = 4 rs = \frac{1}{3}$$

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

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}

$$(2 - u) (2 + u) = \frac{1}{3}$$

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

$$4 - u^2 = \frac{1}{3}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{1}{3}-4 = -\frac{11}{3}$$

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

$$u^2 = \frac{11}{3} u = \pm\sqrt{\frac{11}{3}} = \pm \frac{\sqrt{11}}{\sqrt{3}}$$

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

$$r =2 - \frac{\sqrt{11}}{\sqrt{3}} = 0.085 s = 2 + \frac{\sqrt{11}}{\sqrt{3}} = 3.915$$

Evaluate

$$-3t^{2}+12t-1$$