Solve -25t^2+20t+15

$$- 25 t ^ { 2 } + 20 t + 15$$

Factor

$$-25\left(t-\frac{2-\sqrt{19}}{5}\right)\left(t-\frac{\sqrt{19}+2}{5}\right)$$

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.

$$-25t^{2}+20t+15=0$$

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.

$$t=\frac{-20±\sqrt{20^{2}-4\left(-25\right)\times 15}}{2\left(-25\right)}$$

Square 20.

$$t=\frac{-20±\sqrt{400-4\left(-25\right)\times 15}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$t=\frac{-20±\sqrt{400+100\times 15}}{2\left(-25\right)}$$

Multiply 100 times 15.

$$t=\frac{-20±\sqrt{400+1500}}{2\left(-25\right)}$$

Add 400 to 1500.

$$t=\frac{-20±\sqrt{1900}}{2\left(-25\right)}$$

Take the square root of 1900.

$$t=\frac{-20±10\sqrt{19}}{2\left(-25\right)}$$

Multiply 2 times -25.

$$t=\frac{-20±10\sqrt{19}}{-50}$$

Now solve the equation t=\frac{-20±10\sqrt{19}}{-50} when ± is plus. Add -20 to 10\sqrt{19}.

$$t=\frac{10\sqrt{19}-20}{-50}$$

Divide -20+10\sqrt{19} by -50.

$$t=\frac{2-\sqrt{19}}{5}$$

Now solve the equation t=\frac{-20±10\sqrt{19}}{-50} when ± is minus. Subtract 10\sqrt{19} from -20.

$$t=\frac{-10\sqrt{19}-20}{-50}$$

Divide -20-10\sqrt{19} by -50.

$$t=\frac{\sqrt{19}+2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2-\sqrt{19}}{5} for x_{1} and \frac{2+\sqrt{19}}{5} for x_{2}.

$$-25t^{2}+20t+15=-25\left(t-\frac{2-\sqrt{19}}{5}\right)\left(t-\frac{\sqrt{19}+2}{5}\right)$$

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.

$$x ^ 2 -\frac{4}{5}x -\frac{3}{5} = 0$$

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

$$r + s = \frac{4}{5} rs = -\frac{3}{5}$$

Two numbers r and s sum up to \frac{4}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{5} = \frac{2}{5}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}

$$(\frac{2}{5} - u) (\frac{2}{5} + u) = -\frac{3}{5}$$

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

$$\frac{4}{25} - u^2 = -\frac{3}{5}$$

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

$$-u^2 = -\frac{3}{5}-\frac{4}{25} = -\frac{19}{25}$$

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

$$u^2 = \frac{19}{25} u = \pm\sqrt{\frac{19}{25}} = \pm \frac{\sqrt{19}}{5}$$

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

$$r =\frac{2}{5} - \frac{\sqrt{19}}{5} = -0.472 s = \frac{2}{5} + \frac{\sqrt{19}}{5} = 1.272$$

Evaluate

$$15+20t-25t^{2}$$