Find the value of c that would make the expression a perfect square. 7y^2-70y+c

To make the expression $7y^2 - 70y + c$ a perfect square, we need to find a value for $c$ such that the expression can be written in the form $(ay - b)^2$.<br /><br />First, factor out the coefficient of the $y^2$ term:<br />$7(y^2 - 10y) + c$<br /><br />Now, we need to complete the square inside the parentheses. We take half of the coefficient of the $y$ term, which is $-10$, and square it:<br />$(-10/2)^2 = (-5)^2 = 25$<br /><br />So, we have:<br />$7(y^2 - 10y + 25) + c - 7(25)$<br />$7(y - 5)^2 + c - 175$<br /><br />For the expression to be a perfect square, the term $c - 175$ must be equal to 0. Therefore,<br />$c - 175 = 0$<br />$c = 175$<br /><br />So, the expression becomes:<br />$7(y - 5)^2 = (\sqrt{7}(y - 5))^2 = (\sqrt{7}y - 5\sqrt{7})^2$<br /><br />Thus, the value of $c$ that makes the expression a perfect square is 175.<br /><br />Final Answer: The final answer is $\boxed{175}$