Factor the expression by grouping. First, the expression needs to be rewritten as $50y^{2}+ay+by-28$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=50\left(-28\right)=-1400$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-1400$.
Rewrite $50y^{2}+5y-28$ as $\left(50y^{2}-35y\right)+\left(40y-28\right)$.
$$\left(50y^{2}-35y\right)+\left(40y-28\right)$$
Factor out $5y$ in the first and $4$ in the second group.
$$5y\left(10y-7\right)+4\left(10y-7\right)$$
Factor out common term $10y-7$ by using distributive property.
$$\left(10y-7\right)\left(5y+4\right)$$
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 $ax^{2}+bx+c=0$.
$$50y^{2}+5y-28=0$$
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 $y=\frac{-5±75}{100}$ when $±$ is plus. Add $-5$ to $75$.
$$y=\frac{70}{100}$$
Reduce the fraction $\frac{70}{100}$ to lowest terms by extracting and canceling out $10$.
$$y=\frac{7}{10}$$
Now solve the equation $y=\frac{-5±75}{100}$ when $±$ is minus. Subtract $75$ from $-5$.
$$y=-\frac{80}{100}$$
Reduce the fraction $\frac{-80}{100}$ to lowest terms by extracting and canceling out $20$.
$$y=-\frac{4}{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{7}{10}$ for $x_{1}$ and $-\frac{4}{5}$ for $x_{2}$.
Multiply $\frac{10y-7}{10}$ times $\frac{5y+4}{5}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
