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 $ax^{2}+bx+c=0$.
$$2\alpha ^{2}-150\alpha +1400=0$$
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.
Now solve the equation $\alpha =\frac{150±10\sqrt{113}}{4}$ when $±$ is plus. Add $150$ to $10\sqrt{113}$.
$$\alpha =\frac{10\sqrt{113}+150}{4}$$
Divide $150+10\sqrt{113}$ by $4$.
$$\alpha =\frac{5\sqrt{113}+75}{2}$$
Now solve the equation $\alpha =\frac{150±10\sqrt{113}}{4}$ when $±$ is minus. Subtract $10\sqrt{113}$ from $150$.
$$\alpha =\frac{150-10\sqrt{113}}{4}$$
Divide $150-10\sqrt{113}$ by $4$.
$$\alpha =\frac{75-5\sqrt{113}}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{75+5\sqrt{113}}{2}$ for $x_{1}$ and $\frac{75-5\sqrt{113}}{2}$ for $x_{2}$.
