Solve 54t^2+106t+1330

$$54 t ^ { 2 } + 106 t + 1330$$

Factor

$$2\left(27t^{2}+53t+665\right)$$

Solution Steps

Factor out 2. Polynomial 27t^{2}+53t+665 is not factored since it does not have any rational roots.

$$2\left(27t^{2}+53t+665\right)$$

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.

$$54t^{2}+106t+1330=0$$

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.

$$t=\frac{-106±\sqrt{106^{2}-4\times 54\times 1330}}{2\times 54}$$

Square 106.

$$t=\frac{-106±\sqrt{11236-4\times 54\times 1330}}{2\times 54}$$

Multiply -4 times 54.

$$t=\frac{-106±\sqrt{11236-216\times 1330}}{2\times 54}$$

Multiply -216 times 1330.

$$t=\frac{-106±\sqrt{11236-287280}}{2\times 54}$$

Add 11236 to -287280.

$$t=\frac{-106±\sqrt{-276044}}{2\times 54}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$54t^{2}+106t+1330$$

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 54

$$x ^ 2 +\frac{53}{27}x +\frac{665}{27} = 0$$

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

$$r + s = -\frac{53}{27} rs = \frac{665}{27}$$

Two numbers r and s sum up to -\frac{53}{27} exactly when the average of the two numbers is \frac{1}{2}*-\frac{53}{27} = -\frac{53}{54}. 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>

$$r = -\frac{53}{54} - u s = -\frac{53}{54} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{665}{27}

$$(-\frac{53}{54} - u) (-\frac{53}{54} + u) = \frac{665}{27}$$

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

$$\frac{2809}{2916} - u^2 = \frac{665}{27}$$

Simplify the expression by subtracting \frac{2809}{2916} on both sides

$$-u^2 = \frac{665}{27}-\frac{2809}{2916} = \frac{69011}{2916}$$

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

$$u^2 = -\frac{69011}{2916} u = \pm\sqrt{-\frac{69011}{2916}} = \pm \frac{\sqrt{69011}}{54}i$$

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

$$r =-\frac{53}{54} - \frac{\sqrt{69011}}{54}i = -0.981 - 4.865i s = -\frac{53}{54} + \frac{\sqrt{69011}}{54}i = -0.981 + 4.865i$$

Evaluate

$$54t^{2}+106t+1330$$