Solve 3t^2-11t+4

$$3 t ^ { 2 } - 11 t + 4$$

Factor

$$3\left(t-\frac{11-\sqrt{73}}{6}\right)\left(t-\frac{\sqrt{73}+11}{6}\right)$$

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.

$$3t^{2}-11t+4=0$$

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.

$$t=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Square -11.

$$t=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-11\right)±\sqrt{121-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$t=\frac{-\left(-11\right)±\sqrt{121-48}}{2\times 3}$$

Add 121 to -48.

$$t=\frac{-\left(-11\right)±\sqrt{73}}{2\times 3}$$

The opposite of -11 is 11.

$$t=\frac{11±\sqrt{73}}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{11±\sqrt{73}}{6}$$

Now solve the equation t=\frac{11±\sqrt{73}}{6} when ± is plus. Add 11 to \sqrt{73}.

$$t=\frac{\sqrt{73}+11}{6}$$

Now solve the equation t=\frac{11±\sqrt{73}}{6} when ± is minus. Subtract \sqrt{73} from 11.

$$t=\frac{11-\sqrt{73}}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{11+\sqrt{73}}{6} for x_{1} and \frac{11-\sqrt{73}}{6} for x_{2}.

$$3t^{2}-11t+4=3\left(t-\frac{\sqrt{73}+11}{6}\right)\left(t-\frac{11-\sqrt{73}}{6}\right)$$

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

$$x ^ 2 -\frac{11}{3}x +\frac{4}{3} = 0$$

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

$$r + s = \frac{11}{3} rs = \frac{4}{3}$$

Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{6}. 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>

$$r = \frac{11}{6} - u s = \frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{3}

$$(\frac{11}{6} - u) (\frac{11}{6} + u) = \frac{4}{3}$$

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

$$\frac{121}{36} - u^2 = \frac{4}{3}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = \frac{4}{3}-\frac{121}{36} = -\frac{73}{36}$$

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

$$u^2 = \frac{73}{36} u = \pm\sqrt{\frac{73}{36}} = \pm \frac{\sqrt{73}}{6}$$

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

$$r =\frac{11}{6} - \frac{\sqrt{73}}{6} = 0.409 s = \frac{11}{6} + \frac{\sqrt{73}}{6} = 3.257$$

Evaluate

$$3t^{2}-11t+4$$