Math Problems

Find x-intercepts of the given quadratic equation

## Questions

 $$y = 3 x^2 + x$$ $$y = x^2 + 9 x + 18$$ $$y = 3 x^2 -3 x -6$$ $$y = 3 x^2$$ $$y = x^2 -11 x + 30$$ $$y = x^2 + 8 x + 16$$ $$y = x^2 + 5 x + 6$$ $$y = x^2 -3 x + 2$$ $$y = 2 x^2 + 2 x$$ $$y = x^2 + 6 x + 8$$ $$y = 2 x^2 + 2 x -4$$ $$y = x^2 -3 x -18$$ $$y = 3 x^2 -6 x + 3$$ $$y = 4 x^2 -5 x + 1$$ $$y = 3 x^2 -8 x -3$$ $$y = x^2 -2 x -3$$ $$y = x^2 -10 x + 24$$ $$y = x^2 -5 x -6$$ $$y = 3 x^2 -6 x$$ $$y = x^2 -2 x -3$$ $$y = x^2 -2 x -8$$ $$y = 4 x^2 + 8 x$$ $$y = x^2 + x -6$$ $$y = x^2 -5 x + 6$$ $$y = 4 x^2 + 3 x$$ $$y = x^2 -2 x -15$$ $$y = x^2 -4 x + 3$$ $$y = x^2 -11 x + 30$$

 $$y = 3 x^2 + x$$ ⇒ $$-1 / 3, 0$$ $$y = x^2 + 9 x + 18$$ ⇒ $$-3, -6$$ $$y = 3 x^2 -3 x -6$$ ⇒ $$-1, 2$$ $$y = 3 x^2$$ ⇒ $$0, 0$$ $$y = x^2 -11 x + 30$$ ⇒ $$5, 6$$ $$y = x^2 + 8 x + 16$$ ⇒ $$-4, -4$$ $$y = x^2 + 5 x + 6$$ ⇒ $$-2, -3$$ $$y = x^2 -3 x + 2$$ ⇒ $$2, 1$$ $$y = 2 x^2 + 2 x$$ ⇒ $$-1, 0$$ $$y = x^2 + 6 x + 8$$ ⇒ $$-4, -2$$ $$y = 2 x^2 + 2 x -4$$ ⇒ $$1, -2$$ $$y = x^2 -3 x -18$$ ⇒ $$-3, 6$$ $$y = 3 x^2 -6 x + 3$$ ⇒ $$1, 1$$ $$y = 4 x^2 -5 x + 1$$ ⇒ $$1 / 4, 1$$ $$y = 3 x^2 -8 x -3$$ ⇒ $$-1 / 3, 3$$ $$y = x^2 -2 x -3$$ ⇒ $$3, -1$$ $$y = x^2 -10 x + 24$$ ⇒ $$4, 6$$ $$y = x^2 -5 x -6$$ ⇒ $$-1, 6$$ $$y = 3 x^2 -6 x$$ ⇒ $$0, 2$$ $$y = x^2 -2 x -3$$ ⇒ $$-1, 3$$ $$y = x^2 -2 x -8$$ ⇒ $$4, -2$$ $$y = 4 x^2 + 8 x$$ ⇒ $$0, -2$$ $$y = x^2 + x -6$$ ⇒ $$2, -3$$ $$y = x^2 -5 x + 6$$ ⇒ $$2, 3$$ $$y = 4 x^2 + 3 x$$ ⇒ $$-3 / 4, 0$$ $$y = x^2 -2 x -15$$ ⇒ $$-3, 5$$ $$y = x^2 -4 x + 3$$ ⇒ $$3, 1$$ $$y = x^2 -11 x + 30$$ ⇒ $$6, 5$$