Solve -20x^2+2500x-4000

$$- 20 x ^ { 2 } + 2500 x - 4000$$

Factor

$$-20\left(x-\frac{125-5\sqrt{593}}{2}\right)\left(x-\frac{5\sqrt{593}+125}{2}\right)$$

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.

$$-20x^{2}+2500x-4000=0$$

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.

$$x=\frac{-2500±\sqrt{2500^{2}-4\left(-20\right)\left(-4000\right)}}{2\left(-20\right)}$$

Square 2500.

$$x=\frac{-2500±\sqrt{6250000-4\left(-20\right)\left(-4000\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-2500±\sqrt{6250000+80\left(-4000\right)}}{2\left(-20\right)}$$

Multiply 80 times -4000.

$$x=\frac{-2500±\sqrt{6250000-320000}}{2\left(-20\right)}$$

Add 6250000 to -320000.

$$x=\frac{-2500±\sqrt{5930000}}{2\left(-20\right)}$$

Take the square root of 5930000.

$$x=\frac{-2500±100\sqrt{593}}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-2500±100\sqrt{593}}{-40}$$

Now solve the equation x=\frac{-2500±100\sqrt{593}}{-40} when ± is plus. Add -2500 to 100\sqrt{593}.

$$x=\frac{100\sqrt{593}-2500}{-40}$$

Divide -2500+100\sqrt{593} by -40.

$$x=\frac{125-5\sqrt{593}}{2}$$

Now solve the equation x=\frac{-2500±100\sqrt{593}}{-40} when ± is minus. Subtract 100\sqrt{593} from -2500.

$$x=\frac{-100\sqrt{593}-2500}{-40}$$

Divide -2500-100\sqrt{593} by -40.

$$x=\frac{5\sqrt{593}+125}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{125-5\sqrt{593}}{2} for x_{1} and \frac{125+5\sqrt{593}}{2} for x_{2}.

$$-20x^{2}+2500x-4000=-20\left(x-\frac{125-5\sqrt{593}}{2}\right)\left(x-\frac{5\sqrt{593}+125}{2}\right)$$

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.

$$x ^ 2 -125x +200 = 0$$

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

$$r + s = 125 rs = 200$$

Two numbers r and s sum up to 125 exactly when the average of the two numbers is \frac{1}{2}*125 = \frac{125}{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>

$$r = \frac{125}{2} - u s = \frac{125}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 200

$$(\frac{125}{2} - u) (\frac{125}{2} + u) = 200$$

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

$$\frac{15625}{4} - u^2 = 200$$

Simplify the expression by subtracting \frac{15625}{4} on both sides

$$-u^2 = 200-\frac{15625}{4} = -\frac{14825}{4}$$

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

$$u^2 = \frac{14825}{4} u = \pm\sqrt{\frac{14825}{4}} = \pm \frac{\sqrt{14825}}{2}$$

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

$$r =\frac{125}{2} - \frac{\sqrt{14825}}{2} = 1.621 s = \frac{125}{2} + \frac{\sqrt{14825}}{2} = 123.379$$

Evaluate

$$-20x^{2}+2500x-4000$$