Solve x^2-10x+c+11-c

$$x ^ { 2 } - 10 x + c + 11 - c$$

Factor

$$\left(x-\left(5-\sqrt{14}\right)\right)\left(x-\left(\sqrt{14}+5\right)\right)$$

Steps Using the Quadratic Formula

Combine c and -c to get 0.

$$factor(x^{2}-10x+11)$$

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.

$$x^{2}-10x+11=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 11}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 11}}{2}$$

Multiply -4 times 11.

$$x=\frac{-\left(-10\right)±\sqrt{100-44}}{2}$$

Add 100 to -44.

$$x=\frac{-\left(-10\right)±\sqrt{56}}{2}$$

Take the square root of 56.

$$x=\frac{-\left(-10\right)±2\sqrt{14}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{14}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{14}}{2} when ± is plus. Add 10 to 2\sqrt{14}.

$$x=\frac{2\sqrt{14}+10}{2}$$

Divide 10+2\sqrt{14} by 2.

$$x=\sqrt{14}+5$$

Now solve the equation x=\frac{10±2\sqrt{14}}{2} when ± is minus. Subtract 2\sqrt{14} from 10.

$$x=\frac{10-2\sqrt{14}}{2}$$

Divide 10-2\sqrt{14} by 2.

$$x=5-\sqrt{14}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5+\sqrt{14} for x_{1} and 5-\sqrt{14} for x_{2}.

$$x^{2}-10x+11=\left(x-\left(\sqrt{14}+5\right)\right)\left(x-\left(5-\sqrt{14}\right)\right)$$

Evaluate

$$x^{2}-10x+11$$

Solution Steps

Combine c and -c to get 0.

$$x^{2}-10x+11$$