Solve x^2+x-126=

$${ x }^{ 2 } +x-126=$$

Factor

$$\left(x-\frac{-\sqrt{505}-1}{2}\right)\left(x-\frac{\sqrt{505}-1}{2}\right)$$

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.

$$x^{2}+x-126=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-126\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-126\right)}}{2}$$

Multiply -4 times -126.

$$x=\frac{-1±\sqrt{1+504}}{2}$$

Add 1 to 504.

$$x=\frac{-1±\sqrt{505}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{505}}{2} when ± is plus. Add -1 to \sqrt{505}.

$$x=\frac{\sqrt{505}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{505}}{2} when ± is minus. Subtract \sqrt{505} from -1.

$$x=\frac{-\sqrt{505}-1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{505}}{2} for x_{1} and \frac{-1-\sqrt{505}}{2} for x_{2}.

$$x^{2}+x-126=\left(x-\frac{\sqrt{505}-1}{2}\right)\left(x-\frac{-\sqrt{505}-1}{2}\right)$$

Evaluate

$$x^{2}+x-126$$