Solve 5x^2-40x-4=

$$5 x ^ { 2 } - 40 x - 4 =$$

Factor

$$5\left(x-\left(-\frac{2\sqrt{105}}{5}+4\right)\right)\left(x-\left(\frac{2\sqrt{105}}{5}+4\right)\right)$$

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.

$$5x^{2}-40x-4=0$$

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.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$x=\frac{-\left(-40\right)±\sqrt{1600+80}}{2\times 5}$$

Add 1600 to 80.

$$x=\frac{-\left(-40\right)±\sqrt{1680}}{2\times 5}$$

Take the square root of 1680.

$$x=\frac{-\left(-40\right)±4\sqrt{105}}{2\times 5}$$

The opposite of -40 is 40.

$$x=\frac{40±4\sqrt{105}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{40±4\sqrt{105}}{10}$$

Now solve the equation x=\frac{40±4\sqrt{105}}{10} when ± is plus. Add 40 to 4\sqrt{105}.

$$x=\frac{4\sqrt{105}+40}{10}$$

Divide 40+4\sqrt{105} by 10.

$$x=\frac{2\sqrt{105}}{5}+4$$

Now solve the equation x=\frac{40±4\sqrt{105}}{10} when ± is minus. Subtract 4\sqrt{105} from 40.

$$x=\frac{40-4\sqrt{105}}{10}$$

Divide 40-4\sqrt{105} by 10.

$$x=-\frac{2\sqrt{105}}{5}+4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4+\frac{2\sqrt{105}}{5} for x_{1} and 4-\frac{2\sqrt{105}}{5} for x_{2}.

$$5x^{2}-40x-4=5\left(x-\left(\frac{2\sqrt{105}}{5}+4\right)\right)\left(x-\left(-\frac{2\sqrt{105}}{5}+4\right)\right)$$

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 5

$$x ^ 2 -8x -\frac{4}{5} = 0$$

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

$$r + s = 8 rs = -\frac{4}{5}$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}

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

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

$$16 - u^2 = -\frac{4}{5}$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -\frac{4}{5}-16 = -\frac{84}{5}$$

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

$$u^2 = \frac{84}{5} u = \pm\sqrt{\frac{84}{5}} = \pm \frac{\sqrt{84}}{\sqrt{5}}$$

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

$$r =4 - \frac{\sqrt{84}}{\sqrt{5}} = -0.099 s = 4 + \frac{\sqrt{84}}{\sqrt{5}} = 8.099$$

Evaluate

$$5x^{2}-40x-4$$