Solve a(t)=12t^2+6t-4

$$a ( t ) = 12 t ^ { 2 } + 6 t - 4$$

Factor

$$12\left(t-\left(-\frac{\sqrt{57}}{12}-\frac{1}{4}\right)\right)\left(t-\left(\frac{\sqrt{57}}{12}-\frac{1}{4}\right)\right)$$

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.

$$12t^{2}+6t-4=0$$

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.

$$t=\frac{-6±\sqrt{6^{2}-4\times 12\left(-4\right)}}{2\times 12}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\times 12\left(-4\right)}}{2\times 12}$$

Multiply -4 times 12.

$$t=\frac{-6±\sqrt{36-48\left(-4\right)}}{2\times 12}$$

Multiply -48 times -4.

$$t=\frac{-6±\sqrt{36+192}}{2\times 12}$$

Add 36 to 192.

$$t=\frac{-6±\sqrt{228}}{2\times 12}$$

Take the square root of 228.

$$t=\frac{-6±2\sqrt{57}}{2\times 12}$$

Multiply 2 times 12.

$$t=\frac{-6±2\sqrt{57}}{24}$$

Now solve the equation t=\frac{-6±2\sqrt{57}}{24} when ± is plus. Add -6 to 2\sqrt{57}.

$$t=\frac{2\sqrt{57}-6}{24}$$

Divide -6+2\sqrt{57} by 24.

$$t=\frac{\sqrt{57}}{12}-\frac{1}{4}$$

Now solve the equation t=\frac{-6±2\sqrt{57}}{24} when ± is minus. Subtract 2\sqrt{57} from -6.

$$t=\frac{-2\sqrt{57}-6}{24}$$

Divide -6-2\sqrt{57} by 24.

$$t=-\frac{\sqrt{57}}{12}-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{4}+\frac{\sqrt{57}}{12} for x_{1} and -\frac{1}{4}-\frac{\sqrt{57}}{12} for x_{2}.

$$12t^{2}+6t-4=12\left(t-\left(\frac{\sqrt{57}}{12}-\frac{1}{4}\right)\right)\left(t-\left(-\frac{\sqrt{57}}{12}-\frac{1}{4}\right)\right)$$

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 12

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

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

$$r + s = -\frac{1}{2} rs = -\frac{1}{3}$$

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

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

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

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{1}{3}$$

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

$$\frac{1}{16} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{1}{3}-\frac{1}{16} = -\frac{19}{48}$$

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

$$u^2 = \frac{19}{48} u = \pm\sqrt{\frac{19}{48}} = \pm \frac{\sqrt{19}}{\sqrt{48}}$$

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

$$r =-\frac{1}{4} - \frac{\sqrt{19}}{\sqrt{48}} = -0.879 s = -\frac{1}{4} + \frac{\sqrt{19}}{\sqrt{48}} = 0.379$$

Evaluate

$$12t^{2}+6t-4$$