Solve x^2+5x-950

$${ x }^{ 2 } +5x-950$$

Factor

$$\left(x-\frac{-15\sqrt{17}-5}{2}\right)\left(x-\frac{15\sqrt{17}-5}{2}\right)$$

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.

$$x^{2}+5x-950=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-950\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-950\right)}}{2}$$

Multiply -4 times -950.

$$x=\frac{-5±\sqrt{25+3800}}{2}$$

Add 25 to 3800.

$$x=\frac{-5±\sqrt{3825}}{2}$$

Take the square root of 3825.

$$x=\frac{-5±15\sqrt{17}}{2}$$

Now solve the equation x=\frac{-5±15\sqrt{17}}{2} when ± is plus. Add -5 to 15\sqrt{17}.

$$x=\frac{15\sqrt{17}-5}{2}$$

Now solve the equation x=\frac{-5±15\sqrt{17}}{2} when ± is minus. Subtract 15\sqrt{17} from -5.

$$x=\frac{-15\sqrt{17}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+15\sqrt{17}}{2} for x_{1} and \frac{-5-15\sqrt{17}}{2} for x_{2}.

$$x^{2}+5x-950=\left(x-\frac{15\sqrt{17}-5}{2}\right)\left(x-\frac{-15\sqrt{17}-5}{2}\right)$$

Evaluate

$$x^{2}+5x-950$$