Solve 5t^2-10t-17520

$$5 t ^ { 2 } - 10 t - 17520$$

Factor

$$5\left(t-\left(1-\sqrt{3505}\right)\right)\left(t-\left(\sqrt{3505}+1\right)\right)$$

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.

$$5t^{2}-10t-17520=0$$

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.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\left(-17520\right)}}{2\times 5}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\times 5\left(-17520\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-\left(-10\right)±\sqrt{100-20\left(-17520\right)}}{2\times 5}$$

Multiply -20 times -17520.

$$t=\frac{-\left(-10\right)±\sqrt{100+350400}}{2\times 5}$$

Add 100 to 350400.

$$t=\frac{-\left(-10\right)±\sqrt{350500}}{2\times 5}$$

Take the square root of 350500.

$$t=\frac{-\left(-10\right)±10\sqrt{3505}}{2\times 5}$$

The opposite of -10 is 10.

$$t=\frac{10±10\sqrt{3505}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{10±10\sqrt{3505}}{10}$$

Now solve the equation t=\frac{10±10\sqrt{3505}}{10} when ± is plus. Add 10 to 10\sqrt{3505}.

$$t=\frac{10\sqrt{3505}+10}{10}$$

Divide 10+10\sqrt{3505} by 10.

$$t=\sqrt{3505}+1$$

Now solve the equation t=\frac{10±10\sqrt{3505}}{10} when ± is minus. Subtract 10\sqrt{3505} from 10.

$$t=\frac{10-10\sqrt{3505}}{10}$$

Divide 10-10\sqrt{3505} by 10.

$$t=1-\sqrt{3505}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1+\sqrt{3505} for x_{1} and 1-\sqrt{3505} for x_{2}.

$$5t^{2}-10t-17520=5\left(t-\left(\sqrt{3505}+1\right)\right)\left(t-\left(1-\sqrt{3505}\right)\right)$$

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 5

$$x ^ 2 -2x -3504 = 0$$

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

$$r + s = 2 rs = -3504$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3504

$$(1 - u) (1 + u) = -3504$$

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

$$1 - u^2 = -3504$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -3504-1 = -3505$$

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

$$u^2 = 3505 u = \pm\sqrt{3505} = \pm \sqrt{3505}$$

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

$$r =1 - \sqrt{3505} = -58.203 s = 1 + \sqrt{3505} = 60.203$$

Evaluate

$$5\left(t^{2}-2t-3504\right)$$