Solve -10x^2+2660x-40

$$- 10 x ^ { 2 } + 2660 x - 40$$

Factor

$$-10\left(x-\left(133-3\sqrt{1965}\right)\right)\left(x-\left(3\sqrt{1965}+133\right)\right)$$

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.

$$-10x^{2}+2660x-40=0$$

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.

$$x=\frac{-2660±\sqrt{2660^{2}-4\left(-10\right)\left(-40\right)}}{2\left(-10\right)}$$

Square 2660.

$$x=\frac{-2660±\sqrt{7075600-4\left(-10\right)\left(-40\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-2660±\sqrt{7075600+40\left(-40\right)}}{2\left(-10\right)}$$

Multiply 40 times -40.

$$x=\frac{-2660±\sqrt{7075600-1600}}{2\left(-10\right)}$$

Add 7075600 to -1600.

$$x=\frac{-2660±\sqrt{7074000}}{2\left(-10\right)}$$

Take the square root of 7074000.

$$x=\frac{-2660±60\sqrt{1965}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-2660±60\sqrt{1965}}{-20}$$

Now solve the equation x=\frac{-2660±60\sqrt{1965}}{-20} when ± is plus. Add -2660 to 60\sqrt{1965}.

$$x=\frac{60\sqrt{1965}-2660}{-20}$$

Divide -2660+60\sqrt{1965} by -20.

$$x=133-3\sqrt{1965}$$

Now solve the equation x=\frac{-2660±60\sqrt{1965}}{-20} when ± is minus. Subtract 60\sqrt{1965} from -2660.

$$x=\frac{-60\sqrt{1965}-2660}{-20}$$

Divide -2660-60\sqrt{1965} by -20.

$$x=3\sqrt{1965}+133$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 133-3\sqrt{1965} for x_{1} and 133+3\sqrt{1965} for x_{2}.

$$-10x^{2}+2660x-40=-10\left(x-\left(133-3\sqrt{1965}\right)\right)\left(x-\left(3\sqrt{1965}+133\right)\right)$$

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.

$$x ^ 2 -266x +4 = 0$$

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

$$r + s = 266 rs = 4$$

Two numbers r and s sum up to 266 exactly when the average of the two numbers is \frac{1}{2}*266 = 133. 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>

$$r = 133 - u s = 133 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 4

$$(133 - u) (133 + u) = 4$$

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

$$17689 - u^2 = 4$$

Simplify the expression by subtracting 17689 on both sides

$$-u^2 = 4-17689 = -17685$$

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

$$u^2 = 17685 u = \pm\sqrt{17685} = \pm \sqrt{17685}$$

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

$$r =133 - \sqrt{17685} = 0.015 s = 133 + \sqrt{17685} = 265.985$$

Evaluate

$$-10x^{2}+2660x-40$$