Solve y^2-20y+10=

$${ y }^{ 2 } -20y+10=$$

Factor

$$\left(y-\left(10-3\sqrt{10}\right)\right)\left(y-\left(3\sqrt{10}+10\right)\right)$$

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.

$$y^{2}-20y+10=0$$

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.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 10}}{2}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 10}}{2}$$

Multiply -4 times 10.

$$y=\frac{-\left(-20\right)±\sqrt{400-40}}{2}$$

Add 400 to -40.

$$y=\frac{-\left(-20\right)±\sqrt{360}}{2}$$

Take the square root of 360.

$$y=\frac{-\left(-20\right)±6\sqrt{10}}{2}$$

The opposite of -20 is 20.

$$y=\frac{20±6\sqrt{10}}{2}$$

Now solve the equation y=\frac{20±6\sqrt{10}}{2} when ± is plus. Add 20 to 6\sqrt{10}.

$$y=\frac{6\sqrt{10}+20}{2}$$

Divide 20+6\sqrt{10} by 2.

$$y=3\sqrt{10}+10$$

Now solve the equation y=\frac{20±6\sqrt{10}}{2} when ± is minus. Subtract 6\sqrt{10} from 20.

$$y=\frac{20-6\sqrt{10}}{2}$$

Divide 20-6\sqrt{10} by 2.

$$y=10-3\sqrt{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10+3\sqrt{10} for x_{1} and 10-3\sqrt{10} for x_{2}.

$$y^{2}-20y+10=\left(y-\left(3\sqrt{10}+10\right)\right)\left(y-\left(10-3\sqrt{10}\right)\right)$$

Evaluate

$$y^{2}-20y+10$$