Solve x^2+25x-12500

$${ x }^{ 2 } +25x-12500$$

Factor

$$\left(x-100\right)\left(x+125\right)$$

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.

$$x^{2}+25x-12500=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-12500\right)}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-12500\right)}}{2}$$

Multiply -4 times -12500.

$$x=\frac{-25±\sqrt{625+50000}}{2}$$

Add 625 to 50000.

$$x=\frac{-25±\sqrt{50625}}{2}$$

Take the square root of 50625.

$$x=\frac{-25±225}{2}$$

Now solve the equation x=\frac{-25±225}{2} when ± is plus. Add -25 to 225.

$$x=\frac{200}{2}$$

Divide 200 by 2.

$$x=100$$

Now solve the equation x=\frac{-25±225}{2} when ± is minus. Subtract 225 from -25.

$$x=-\frac{250}{2}$$

Divide -250 by 2.

$$x=-125$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 100 for x_{1} and -125 for x_{2}.

$$x^{2}+25x-12500=\left(x-100\right)\left(x-\left(-125\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}+25x-12500=\left(x-100\right)\left(x+125\right)$$

Evaluate

$$\left(x-100\right)\left(x+125\right)$$