Solve x^2+100x-560

$$x ^ { 2 } + 100 x - 560$$

Factor

$$\left(x-\left(-6\sqrt{85}-50\right)\right)\left(x-\left(6\sqrt{85}-50\right)\right)$$

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.

$$x^{2}+100x-560=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-560\right)}}{2}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-560\right)}}{2}$$

Multiply -4 times -560.

$$x=\frac{-100±\sqrt{10000+2240}}{2}$$

Add 10000 to 2240.

$$x=\frac{-100±\sqrt{12240}}{2}$$

Take the square root of 12240.

$$x=\frac{-100±12\sqrt{85}}{2}$$

Now solve the equation x=\frac{-100±12\sqrt{85}}{2} when ± is plus. Add -100 to 12\sqrt{85}.

$$x=\frac{12\sqrt{85}-100}{2}$$

Divide -100+12\sqrt{85} by 2.

$$x=6\sqrt{85}-50$$

Now solve the equation x=\frac{-100±12\sqrt{85}}{2} when ± is minus. Subtract 12\sqrt{85} from -100.

$$x=\frac{-12\sqrt{85}-100}{2}$$

Divide -100-12\sqrt{85} by 2.

$$x=-6\sqrt{85}-50$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -50+6\sqrt{85} for x_{1} and -50-6\sqrt{85} for x_{2}.

$$x^{2}+100x-560=\left(x-\left(6\sqrt{85}-50\right)\right)\left(x-\left(-6\sqrt{85}-50\right)\right)$$

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.

$$x ^ 2 +100x -560 = 0$$

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

$$r + s = -100 rs = -560$$

Two numbers r and s sum up to -100 exactly when the average of the two numbers is \frac{1}{2}*-100 = -50. 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>

$$r = -50 - u s = -50 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -560

$$(-50 - u) (-50 + u) = -560$$

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

$$2500 - u^2 = -560$$

Simplify the expression by subtracting 2500 on both sides

$$-u^2 = -560-2500 = -3060$$

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

$$u^2 = 3060 u = \pm\sqrt{3060} = \pm \sqrt{3060}$$

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

$$r =-50 - \sqrt{3060} = -105.317 s = -50 + \sqrt{3060} = 5.317$$

Evaluate

$$x^{2}+100x-560$$