Solve -6x^2+400x-4000

$$- 6 x ^ { 2 } + 400 x - 4000$$

Factor

$$-6\left(x-\frac{100-20\sqrt{10}}{3}\right)\left(x-\frac{20\sqrt{10}+100}{3}\right)$$

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.

$$-6x^{2}+400x-4000=0$$

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.

$$x=\frac{-400±\sqrt{400^{2}-4\left(-6\right)\left(-4000\right)}}{2\left(-6\right)}$$

Square 400.

$$x=\frac{-400±\sqrt{160000-4\left(-6\right)\left(-4000\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-400±\sqrt{160000+24\left(-4000\right)}}{2\left(-6\right)}$$

Multiply 24 times -4000.

$$x=\frac{-400±\sqrt{160000-96000}}{2\left(-6\right)}$$

Add 160000 to -96000.

$$x=\frac{-400±\sqrt{64000}}{2\left(-6\right)}$$

Take the square root of 64000.

$$x=\frac{-400±80\sqrt{10}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-400±80\sqrt{10}}{-12}$$

Now solve the equation x=\frac{-400±80\sqrt{10}}{-12} when ± is plus. Add -400 to 80\sqrt{10}.

$$x=\frac{80\sqrt{10}-400}{-12}$$

Divide -400+80\sqrt{10} by -12.

$$x=\frac{100-20\sqrt{10}}{3}$$

Now solve the equation x=\frac{-400±80\sqrt{10}}{-12} when ± is minus. Subtract 80\sqrt{10} from -400.

$$x=\frac{-80\sqrt{10}-400}{-12}$$

Divide -400-80\sqrt{10} by -12.

$$x=\frac{20\sqrt{10}+100}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{100-20\sqrt{10}}{3} for x_{1} and \frac{100+20\sqrt{10}}{3} for x_{2}.

$$-6x^{2}+400x-4000=-6\left(x-\frac{100-20\sqrt{10}}{3}\right)\left(x-\frac{20\sqrt{10}+100}{3}\right)$$

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.

$$x ^ 2 -\frac{200}{3}x +\frac{2000}{3} = 0$$

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

$$r + s = \frac{200}{3} rs = \frac{2000}{3}$$

Two numbers r and s sum up to \frac{200}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{200}{3} = \frac{100}{3}. 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>

$$r = \frac{100}{3} - u s = \frac{100}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2000}{3}

$$(\frac{100}{3} - u) (\frac{100}{3} + u) = \frac{2000}{3}$$

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

$$\frac{10000}{9} - u^2 = \frac{2000}{3}$$

Simplify the expression by subtracting \frac{10000}{9} on both sides

$$-u^2 = \frac{2000}{3}-\frac{10000}{9} = -\frac{4000}{9}$$

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

$$u^2 = \frac{4000}{9} u = \pm\sqrt{\frac{4000}{9}} = \pm \frac{\sqrt{4000}}{3}$$

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

$$r =\frac{100}{3} - \frac{\sqrt{4000}}{3} = 12.251 s = \frac{100}{3} + \frac{\sqrt{4000}}{3} = 54.415$$

Evaluate

$$-6x^{2}+400x-4000$$