Solve x^2-10x-3

$${ x }^{ 2 } -10x-3$$

Factor

$$\left(x-\left(5-2\sqrt{7}\right)\right)\left(x-\left(2\sqrt{7}+5\right)\right)$$

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.

$$x^{2}-10x-3=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12}}{2}$$

Add 100 to 12.

$$x=\frac{-\left(-10\right)±\sqrt{112}}{2}$$

Take the square root of 112.

$$x=\frac{-\left(-10\right)±4\sqrt{7}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±4\sqrt{7}}{2}$$

Now solve the equation x=\frac{10±4\sqrt{7}}{2} when ± is plus. Add 10 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}+10}{2}$$

Divide 10+4\sqrt{7} by 2.

$$x=2\sqrt{7}+5$$

Now solve the equation x=\frac{10±4\sqrt{7}}{2} when ± is minus. Subtract 4\sqrt{7} from 10.

$$x=\frac{10-4\sqrt{7}}{2}$$

Divide 10-4\sqrt{7} by 2.

$$x=5-2\sqrt{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5+2\sqrt{7} for x_{1} and 5-2\sqrt{7} for x_{2}.

$$x^{2}-10x-3=\left(x-\left(2\sqrt{7}+5\right)\right)\left(x-\left(5-2\sqrt{7}\right)\right)$$

Evaluate

$$x^{2}-10x-3$$