Solve f(x)=25x^2+300x+894

$$f ( x ) = 25 x ^ { 2 } + 300 x + 894$$

Factor

$$25\left(x-\left(-\frac{\sqrt{6}}{5}-6\right)\right)\left(x-\left(\frac{\sqrt{6}}{5}-6\right)\right)$$

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.

$$25x^{2}+300x+894=0$$

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.

$$x=\frac{-300±\sqrt{300^{2}-4\times 25\times 894}}{2\times 25}$$

Square 300.

$$x=\frac{-300±\sqrt{90000-4\times 25\times 894}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-300±\sqrt{90000-100\times 894}}{2\times 25}$$

Multiply -100 times 894.

$$x=\frac{-300±\sqrt{90000-89400}}{2\times 25}$$

Add 90000 to -89400.

$$x=\frac{-300±\sqrt{600}}{2\times 25}$$

Take the square root of 600.

$$x=\frac{-300±10\sqrt{6}}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-300±10\sqrt{6}}{50}$$

Now solve the equation x=\frac{-300±10\sqrt{6}}{50} when ± is plus. Add -300 to 10\sqrt{6}.

$$x=\frac{10\sqrt{6}-300}{50}$$

Divide -300+10\sqrt{6} by 50.

$$x=\frac{\sqrt{6}}{5}-6$$

Now solve the equation x=\frac{-300±10\sqrt{6}}{50} when ± is minus. Subtract 10\sqrt{6} from -300.

$$x=\frac{-10\sqrt{6}-300}{50}$$

Divide -300-10\sqrt{6} by 50.

$$x=-\frac{\sqrt{6}}{5}-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6+\frac{\sqrt{6}}{5} for x_{1} and -6-\frac{\sqrt{6}}{5} for x_{2}.

$$25x^{2}+300x+894=25\left(x-\left(\frac{\sqrt{6}}{5}-6\right)\right)\left(x-\left(-\frac{\sqrt{6}}{5}-6\right)\right)$$

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 25

$$x ^ 2 +12x +\frac{894}{25} = 0$$

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

$$r + s = -12 rs = \frac{894}{25}$$

Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>

$$r = -6 - u s = -6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{894}{25}

$$(-6 - u) (-6 + u) = \frac{894}{25}$$

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

$$36 - u^2 = \frac{894}{25}$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = \frac{894}{25}-36 = -\frac{6}{25}$$

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

$$u^2 = \frac{6}{25} u = \pm\sqrt{\frac{6}{25}} = \pm \frac{\sqrt{6}}{5}$$

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

$$r =-6 - \frac{\sqrt{6}}{5} = -6.490 s = -6 + \frac{\sqrt{6}}{5} = -5.510$$

Evaluate

$$25x^{2}+300x+894$$