Solve x(y)=1+10y-5y^2

$$x(y)=1+10y-5 { y }^{ 2 }$$

Factor

$$-5\left(y-\left(-\frac{\sqrt{30}}{5}+1\right)\right)\left(y-\left(\frac{\sqrt{30}}{5}+1\right)\right)$$

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.

$$-5y^{2}+10y+1=0$$

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.

$$y=\frac{-10±\sqrt{10^{2}-4\left(-5\right)}}{2\left(-5\right)}$$

Square 10.

$$y=\frac{-10±\sqrt{100-4\left(-5\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-10±\sqrt{100+20}}{2\left(-5\right)}$$

Add 100 to 20.

$$y=\frac{-10±\sqrt{120}}{2\left(-5\right)}$$

Take the square root of 120.

$$y=\frac{-10±2\sqrt{30}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{-10±2\sqrt{30}}{-10}$$

Now solve the equation y=\frac{-10±2\sqrt{30}}{-10} when ± is plus. Add -10 to 2\sqrt{30}.

$$y=\frac{2\sqrt{30}-10}{-10}$$

Divide -10+2\sqrt{30} by -10.

$$y=-\frac{\sqrt{30}}{5}+1$$

Now solve the equation y=\frac{-10±2\sqrt{30}}{-10} when ± is minus. Subtract 2\sqrt{30} from -10.

$$y=\frac{-2\sqrt{30}-10}{-10}$$

Divide -10-2\sqrt{30} by -10.

$$y=\frac{\sqrt{30}}{5}+1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1-\frac{\sqrt{30}}{5} for x_{1} and 1+\frac{\sqrt{30}}{5} for x_{2}.

$$-5y^{2}+10y+1=-5\left(y-\left(-\frac{\sqrt{30}}{5}+1\right)\right)\left(y-\left(\frac{\sqrt{30}}{5}+1\right)\right)$$

Evaluate

$$1+10y-5y^{2}$$