Solve 4x^2+1020x-5=

$$4 x ^ { 2 } + 1020 x - 5 =$$

Factor

$$4\left(x-\frac{-\sqrt{65030}-255}{2}\right)\left(x-\frac{\sqrt{65030}-255}{2}\right)$$

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.

$$4x^{2}+1020x-5=0$$

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.

$$x=\frac{-1020±\sqrt{1020^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square 1020.

$$x=\frac{-1020±\sqrt{1040400-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1020±\sqrt{1040400-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$x=\frac{-1020±\sqrt{1040400+80}}{2\times 4}$$

Add 1040400 to 80.

$$x=\frac{-1020±\sqrt{1040480}}{2\times 4}$$

Take the square root of 1040480.

$$x=\frac{-1020±4\sqrt{65030}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1020±4\sqrt{65030}}{8}$$

Now solve the equation x=\frac{-1020±4\sqrt{65030}}{8} when ± is plus. Add -1020 to 4\sqrt{65030}.

$$x=\frac{4\sqrt{65030}-1020}{8}$$

Divide -1020+4\sqrt{65030} by 8.

$$x=\frac{\sqrt{65030}-255}{2}$$

Now solve the equation x=\frac{-1020±4\sqrt{65030}}{8} when ± is minus. Subtract 4\sqrt{65030} from -1020.

$$x=\frac{-4\sqrt{65030}-1020}{8}$$

Divide -1020-4\sqrt{65030} by 8.

$$x=\frac{-\sqrt{65030}-255}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-255+\sqrt{65030}}{2} for x_{1} and \frac{-255-\sqrt{65030}}{2} for x_{2}.

$$4x^{2}+1020x-5=4\left(x-\frac{\sqrt{65030}-255}{2}\right)\left(x-\frac{-\sqrt{65030}-255}{2}\right)$$

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 4

$$x ^ 2 +255x -\frac{5}{4} = 0$$

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

$$r + s = -255 rs = -\frac{5}{4}$$

Two numbers r and s sum up to -255 exactly when the average of the two numbers is \frac{1}{2}*-255 = -\frac{255}{2}. 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>

$$r = -\frac{255}{2} - u s = -\frac{255}{2} + u$$

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

$$(-\frac{255}{2} - u) (-\frac{255}{2} + u) = -\frac{5}{4}$$

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

$$\frac{65025}{4} - u^2 = -\frac{5}{4}$$

Simplify the expression by subtracting \frac{65025}{4} on both sides

$$-u^2 = -\frac{5}{4}-\frac{65025}{4} = -\frac{32515}{2}$$

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

$$u^2 = \frac{32515}{2} u = \pm\sqrt{\frac{32515}{2}} = \pm \frac{\sqrt{32515}}{\sqrt{2}}$$

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

$$r =-\frac{255}{2} - \frac{\sqrt{32515}}{\sqrt{2}} = -255.005 s = -\frac{255}{2} + \frac{\sqrt{32515}}{\sqrt{2}} = 0.005$$

Evaluate

$$4x^{2}+1020x-5$$