Solve 25t^2-45t+9

$$25 t ^ { 2 } - 45 t + 9$$

Factor

$$25\left(t-\frac{9-3\sqrt{5}}{10}\right)\left(t-\frac{3\sqrt{5}+9}{10}\right)$$

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.

$$25t^{2}-45t+9=0$$

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.

$$t=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}-4\times 25\times 9}}{2\times 25}$$

Square -45.

$$t=\frac{-\left(-45\right)±\sqrt{2025-4\times 25\times 9}}{2\times 25}$$

Multiply -4 times 25.

$$t=\frac{-\left(-45\right)±\sqrt{2025-100\times 9}}{2\times 25}$$

Multiply -100 times 9.

$$t=\frac{-\left(-45\right)±\sqrt{2025-900}}{2\times 25}$$

Add 2025 to -900.

$$t=\frac{-\left(-45\right)±\sqrt{1125}}{2\times 25}$$

Take the square root of 1125.

$$t=\frac{-\left(-45\right)±15\sqrt{5}}{2\times 25}$$

The opposite of -45 is 45.

$$t=\frac{45±15\sqrt{5}}{2\times 25}$$

Multiply 2 times 25.

$$t=\frac{45±15\sqrt{5}}{50}$$

Now solve the equation t=\frac{45±15\sqrt{5}}{50} when ± is plus. Add 45 to 15\sqrt{5}.

$$t=\frac{15\sqrt{5}+45}{50}$$

Divide 45+15\sqrt{5} by 50.

$$t=\frac{3\sqrt{5}+9}{10}$$

Now solve the equation t=\frac{45±15\sqrt{5}}{50} when ± is minus. Subtract 15\sqrt{5} from 45.

$$t=\frac{45-15\sqrt{5}}{50}$$

Divide 45-15\sqrt{5} by 50.

$$t=\frac{9-3\sqrt{5}}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9+3\sqrt{5}}{10} for x_{1} and \frac{9-3\sqrt{5}}{10} for x_{2}.

$$25t^{2}-45t+9=25\left(t-\frac{3\sqrt{5}+9}{10}\right)\left(t-\frac{9-3\sqrt{5}}{10}\right)$$

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 25

$$x ^ 2 -\frac{9}{5}x +\frac{9}{25} = 0$$

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

$$r + s = \frac{9}{5} rs = \frac{9}{25}$$

Two numbers r and s sum up to \frac{9}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{5} = \frac{9}{10}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{9}{10} - u s = \frac{9}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{25}

$$(\frac{9}{10} - u) (\frac{9}{10} + u) = \frac{9}{25}$$

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

$$\frac{81}{100} - u^2 = \frac{9}{25}$$

Simplify the expression by subtracting \frac{81}{100} on both sides

$$-u^2 = \frac{9}{25}-\frac{81}{100} = -\frac{9}{20}$$

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

$$u^2 = \frac{9}{20} u = \pm\sqrt{\frac{9}{20}} = \pm \frac{3}{\sqrt{20}}$$

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

$$r =\frac{9}{10} - \frac{3}{\sqrt{20}} = 0.229 s = \frac{9}{10} + \frac{3}{\sqrt{20}} = 1.571$$

Evaluate

$$25t^{2}-45t+9$$