Solve h(t)=-3t^2+24t+1

$$h ( t ) = - 3 t ^ { 2 } + 24 t + 1$$

Factor

$$-3\left(t-\left(-\frac{7\sqrt{3}}{3}+4\right)\right)\left(t-\left(\frac{7\sqrt{3}}{3}+4\right)\right)$$

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.

$$-3t^{2}+24t+1=0$$

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.

$$t=\frac{-24±\sqrt{24^{2}-4\left(-3\right)}}{2\left(-3\right)}$$

Square 24.

$$t=\frac{-24±\sqrt{576-4\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$t=\frac{-24±\sqrt{576+12}}{2\left(-3\right)}$$

Add 576 to 12.

$$t=\frac{-24±\sqrt{588}}{2\left(-3\right)}$$

Take the square root of 588.

$$t=\frac{-24±14\sqrt{3}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$t=\frac{-24±14\sqrt{3}}{-6}$$

Now solve the equation t=\frac{-24±14\sqrt{3}}{-6} when ± is plus. Add -24 to 14\sqrt{3}.

$$t=\frac{14\sqrt{3}-24}{-6}$$

Divide -24+14\sqrt{3} by -6.

$$t=-\frac{7\sqrt{3}}{3}+4$$

Now solve the equation t=\frac{-24±14\sqrt{3}}{-6} when ± is minus. Subtract 14\sqrt{3} from -24.

$$t=\frac{-14\sqrt{3}-24}{-6}$$

Divide -24-14\sqrt{3} by -6.

$$t=\frac{7\sqrt{3}}{3}+4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4-\frac{7\sqrt{3}}{3} for x_{1} and 4+\frac{7\sqrt{3}}{3} for x_{2}.

$$-3t^{2}+24t+1=-3\left(t-\left(-\frac{7\sqrt{3}}{3}+4\right)\right)\left(t-\left(\frac{7\sqrt{3}}{3}+4\right)\right)$$

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.

$$x ^ 2 -8x -\frac{1}{3} = 0$$

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

$$r + s = 8 rs = -\frac{1}{3}$$

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

$$r = 4 - u s = 4 + u$$

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

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

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

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

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -\frac{1}{3}-16 = -\frac{49}{3}$$

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

$$u^2 = \frac{49}{3} u = \pm\sqrt{\frac{49}{3}} = \pm \frac{7}{\sqrt{3}}$$

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

$$r =4 - \frac{7}{\sqrt{3}} = -0.041 s = 4 + \frac{7}{\sqrt{3}} = 8.041$$

Evaluate

$$1+24t-3t^{2}$$