Solve -t^2-5t+1

$$- t ^ { 2 } - 5 t + 1$$

Factor

$$-\left(t-\frac{-\sqrt{29}-5}{2}\right)\left(t-\frac{\sqrt{29}-5}{2}\right)$$

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.

$$-t^{2}-5t+1=0$$

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.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square -5.

$$t=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-5\right)±\sqrt{25+4}}{2\left(-1\right)}$$

Add 25 to 4.

$$t=\frac{-\left(-5\right)±\sqrt{29}}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$t=\frac{5±\sqrt{29}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{5±\sqrt{29}}{-2}$$

Now solve the equation t=\frac{5±\sqrt{29}}{-2} when ± is plus. Add 5 to \sqrt{29}.

$$t=\frac{\sqrt{29}+5}{-2}$$

Divide 5+\sqrt{29} by -2.

$$t=\frac{-\sqrt{29}-5}{2}$$

Now solve the equation t=\frac{5±\sqrt{29}}{-2} when ± is minus. Subtract \sqrt{29} from 5.

$$t=\frac{5-\sqrt{29}}{-2}$$

Divide 5-\sqrt{29} by -2.

$$t=\frac{\sqrt{29}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5-\sqrt{29}}{2} for x_{1} and \frac{-5+\sqrt{29}}{2} for x_{2}.

$$-t^{2}-5t+1=-\left(t-\frac{-\sqrt{29}-5}{2}\right)\left(t-\frac{\sqrt{29}-5}{2}\right)$$

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.

$$x ^ 2 +5x -1 = 0$$

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

$$r + s = -5 rs = -1$$

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{2}. 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>

$$r = -\frac{5}{2} - u s = -\frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -1$$

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

$$\frac{25}{4} - u^2 = -1$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -1-\frac{25}{4} = -\frac{29}{4}$$

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

$$u^2 = \frac{29}{4} u = \pm\sqrt{\frac{29}{4}} = \pm \frac{\sqrt{29}}{2}$$

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

$$r =-\frac{5}{2} - \frac{\sqrt{29}}{2} = -5.193 s = -\frac{5}{2} + \frac{\sqrt{29}}{2} = 0.193$$

Evaluate

$$1-5t-t^{2}$$