Solve x^2+100x-300000

$${ x }^{ 2 } +100x-300000$$

Factor

$$\left(x-500\right)\left(x+600\right)$$

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.

$$x^{2}+100x-300000=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-300000\right)}}{2}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-300000\right)}}{2}$$

Multiply -4 times -300000.

$$x=\frac{-100±\sqrt{10000+1200000}}{2}$$

Add 10000 to 1200000.

$$x=\frac{-100±\sqrt{1210000}}{2}$$

Take the square root of 1210000.

$$x=\frac{-100±1100}{2}$$

Now solve the equation x=\frac{-100±1100}{2} when ± is plus. Add -100 to 1100.

$$x=\frac{1000}{2}$$

Divide 1000 by 2.

$$x=500$$

Now solve the equation x=\frac{-100±1100}{2} when ± is minus. Subtract 1100 from -100.

$$x=-\frac{1200}{2}$$

Divide -1200 by 2.

$$x=-600$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 500 for x_{1} and -600 for x_{2}.

$$x^{2}+100x-300000=\left(x-500\right)\left(x-\left(-600\right)\right)$$

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

$$x^{2}+100x-300000=\left(x-500\right)\left(x+600\right)$$

Evaluate

$$\left(x-500\right)\left(x+600\right)$$