Solve 25x^2+10x-1=

$$25 x ^ { 2 } + 10 x - 1 =$$

Factor

$$25\left(x-\frac{-\sqrt{2}-1}{5}\right)\left(x-\frac{\sqrt{2}-1}{5}\right)$$

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.

$$25x^{2}+10x-1=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 25\left(-1\right)}}{2\times 25}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 25\left(-1\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-10±\sqrt{100-100\left(-1\right)}}{2\times 25}$$

Multiply -100 times -1.

$$x=\frac{-10±\sqrt{100+100}}{2\times 25}$$

Add 100 to 100.

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

Take the square root of 200.

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

Multiply 2 times 25.

$$x=\frac{-10±10\sqrt{2}}{50}$$

Now solve the equation x=\frac{-10±10\sqrt{2}}{50} when ± is plus. Add -10 to 10\sqrt{2}.

$$x=\frac{10\sqrt{2}-10}{50}$$

Divide -10+10\sqrt{2} by 50.

$$x=\frac{\sqrt{2}-1}{5}$$

Now solve the equation x=\frac{-10±10\sqrt{2}}{50} when ± is minus. Subtract 10\sqrt{2} from -10.

$$x=\frac{-10\sqrt{2}-10}{50}$$

Divide -10-10\sqrt{2} by 50.

$$x=\frac{-\sqrt{2}-1}{5}$$

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

$$25x^{2}+10x-1=25\left(x-\frac{\sqrt{2}-1}{5}\right)\left(x-\frac{-\sqrt{2}-1}{5}\right)$$

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.This is achieved by dividing both sides of the equation by 25

$$x ^ 2 +\frac{2}{5}x -\frac{1}{25} = 0$$

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

$$r + s = -\frac{2}{5} rs = -\frac{1}{25}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{25}

$$(-\frac{1}{5} - u) (-\frac{1}{5} + u) = -\frac{1}{25}$$

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

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

Simplify the expression by subtracting \frac{1}{25} on both sides

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

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

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

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

$$r =-\frac{1}{5} - \frac{\sqrt{2}}{5} = -0.483 s = -\frac{1}{5} + \frac{\sqrt{2}}{5} = 0.083$$

Evaluate

$$25x^{2}+10x-1$$