Solve 10x^2+25x+5

$$10 x ^ { 2 } + 25 x + 5$$

Factor

$$10\left(x-\frac{-\sqrt{17}-5}{4}\right)\left(x-\frac{\sqrt{17}-5}{4}\right)$$

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.

$$10x^{2}+25x+5=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 10\times 5}}{2\times 10}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 10\times 5}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-25±\sqrt{625-40\times 5}}{2\times 10}$$

Multiply -40 times 5.

$$x=\frac{-25±\sqrt{625-200}}{2\times 10}$$

Add 625 to -200.

$$x=\frac{-25±\sqrt{425}}{2\times 10}$$

Take the square root of 425.

$$x=\frac{-25±5\sqrt{17}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-25±5\sqrt{17}}{20}$$

Now solve the equation x=\frac{-25±5\sqrt{17}}{20} when ± is plus. Add -25 to 5\sqrt{17}.

$$x=\frac{5\sqrt{17}-25}{20}$$

Divide -25+5\sqrt{17} by 20.

$$x=\frac{\sqrt{17}-5}{4}$$

Now solve the equation x=\frac{-25±5\sqrt{17}}{20} when ± is minus. Subtract 5\sqrt{17} from -25.

$$x=\frac{-5\sqrt{17}-25}{20}$$

Divide -25-5\sqrt{17} by 20.

$$x=\frac{-\sqrt{17}-5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{17}}{4} for x_{1} and \frac{-5-\sqrt{17}}{4} for x_{2}.

$$10x^{2}+25x+5=10\left(x-\frac{\sqrt{17}-5}{4}\right)\left(x-\frac{-\sqrt{17}-5}{4}\right)$$

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 10

$$x ^ 2 +\frac{5}{2}x +\frac{1}{2} = 0$$

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

$$r + s = -\frac{5}{2} rs = \frac{1}{2}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = \frac{1}{2}$$

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

$$\frac{25}{16} - u^2 = \frac{1}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = \frac{1}{2}-\frac{25}{16} = -\frac{17}{16}$$

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

$$u^2 = \frac{17}{16} u = \pm\sqrt{\frac{17}{16}} = \pm \frac{\sqrt{17}}{4}$$

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

$$r =-\frac{5}{4} - \frac{\sqrt{17}}{4} = -2.281 s = -\frac{5}{4} + \frac{\sqrt{17}}{4} = -0.219$$

Evaluate

$$10x^{2}+25x+5$$