### Home > INT3 > Chapter 2 > Lesson 2.1.1 > Problem2-6

2-6.

Solve each of the following equations using a method other than the Quadratic Formula.

1. $y^2 - 6y = 0$

Factor $y$ out of these terms.

$y=0$ or $y=6$

1. $n^2 + 5n + 7 = 7$

Subtract $7$ from both sides, then factor $n$ out of the terms.

$n=0$ or $n=-5$

1. $2t^2 – 14t + 3 = 3$

Refer to part (b).

1. $\frac {1}{3}x^2 + 3x - 4 = -4$

Refer to part (b).

1. Zero is a solution to each of the above equations. What do all of the above equations have in common that causes them to have zero as a solution?

Look at the term without a variable in each equation. Is there a similarity between the constants and the right side of the equation?