Solve 5t^2-30t-25

$$5 t ^ { 2 } - 30 t - 25$$

Factor

$$5\left(t-\left(3-\sqrt{14}\right)\right)\left(t-\left(\sqrt{14}+3\right)\right)$$

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.

$$5t^{2}-30t-25=0$$

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.

$$t=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 5\left(-25\right)}}{2\times 5}$$

Square -30.

$$t=\frac{-\left(-30\right)±\sqrt{900-4\times 5\left(-25\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-\left(-30\right)±\sqrt{900-20\left(-25\right)}}{2\times 5}$$

Multiply -20 times -25.

$$t=\frac{-\left(-30\right)±\sqrt{900+500}}{2\times 5}$$

Add 900 to 500.

$$t=\frac{-\left(-30\right)±\sqrt{1400}}{2\times 5}$$

Take the square root of 1400.

$$t=\frac{-\left(-30\right)±10\sqrt{14}}{2\times 5}$$

The opposite of -30 is 30.

$$t=\frac{30±10\sqrt{14}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{30±10\sqrt{14}}{10}$$

Now solve the equation t=\frac{30±10\sqrt{14}}{10} when ± is plus. Add 30 to 10\sqrt{14}.

$$t=\frac{10\sqrt{14}+30}{10}$$

Divide 30+10\sqrt{14} by 10.

$$t=\sqrt{14}+3$$

Now solve the equation t=\frac{30±10\sqrt{14}}{10} when ± is minus. Subtract 10\sqrt{14} from 30.

$$t=\frac{30-10\sqrt{14}}{10}$$

Divide 30-10\sqrt{14} by 10.

$$t=3-\sqrt{14}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3+\sqrt{14} for x_{1} and 3-\sqrt{14} for x_{2}.

$$5t^{2}-30t-25=5\left(t-\left(\sqrt{14}+3\right)\right)\left(t-\left(3-\sqrt{14}\right)\right)$$

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

$$x ^ 2 -6x -5 = 0$$

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

$$r + s = 6 rs = -5$$

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

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(3 - u) (3 + u) = -5$$

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

$$9 - u^2 = -5$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -5-9 = -14$$

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

$$u^2 = 14 u = \pm\sqrt{14} = \pm \sqrt{14}$$

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

$$r =3 - \sqrt{14} = -0.742 s = 3 + \sqrt{14} = 6.742$$

Evaluate

$$5\left(t^{2}-6t-5\right)$$