Math Problems

Find x-intercepts of the given quadratic equation

Questions

 $$y = x^2 -x -30$$ $$y = x^2 -5 x + 4$$ $$y = x^2 -4 x + 3$$ $$y = x^2 -5 x + 4$$ $$y = 4 x^2 -3 x -1$$ $$y = x^2 -8 x + 12$$ $$y = x^2 -7 x + 10$$ $$y = 4 x^2 + 2 x -2$$ $$y = x^2 + 3 x -10$$ $$y = x^2 + x -6$$ $$y = x^2 -x -12$$ $$y = 3 x^2 + 9 x$$ $$y = x^2 + 6 x + 5$$ $$y = 3 x^2 + 7 x -6$$ $$y = x^2 -4 x -5$$ $$y = x^2 + 3 x -18$$ $$y = x^2 -7 x + 6$$ $$y = 2 x^2 -9 x + 9$$ $$y = 4 x^2 -14 x + 6$$ $$y = x^2 -4 x -12$$ $$y = 4 x^2 -10 x -6$$ $$y = 3 x^2 + 3 x -6$$ $$y = x^2 + 7 x + 12$$ $$y = x^2 + 4 x -5$$ $$y = 2 x^2 -5 x -3$$ $$y = 3 x^2 + 8 x -3$$ $$y = 4 x^2 -9 x -9$$ $$y = x^2 + 8 x + 12$$

 $$y = x^2 -x -30$$ ⇒ $$6, -5$$ $$y = x^2 -5 x + 4$$ ⇒ $$1, 4$$ $$y = x^2 -4 x + 3$$ ⇒ $$1, 3$$ $$y = x^2 -5 x + 4$$ ⇒ $$4, 1$$ $$y = 4 x^2 -3 x -1$$ ⇒ $$-1 / 4, 1$$ $$y = x^2 -8 x + 12$$ ⇒ $$6, 2$$ $$y = x^2 -7 x + 10$$ ⇒ $$2, 5$$ $$y = 4 x^2 + 2 x -2$$ ⇒ $$1 / 2, -1$$ $$y = x^2 + 3 x -10$$ ⇒ $$2, -5$$ $$y = x^2 + x -6$$ ⇒ $$2, -3$$ $$y = x^2 -x -12$$ ⇒ $$4, -3$$ $$y = 3 x^2 + 9 x$$ ⇒ $$0, -3$$ $$y = x^2 + 6 x + 5$$ ⇒ $$-1, -5$$ $$y = 3 x^2 + 7 x -6$$ ⇒ $$2 / 3, -3$$ $$y = x^2 -4 x -5$$ ⇒ $$-1, 5$$ $$y = x^2 + 3 x -18$$ ⇒ $$-6, 3$$ $$y = x^2 -7 x + 6$$ ⇒ $$6, 1$$ $$y = 2 x^2 -9 x + 9$$ ⇒ $$3 / 2, 3$$ $$y = 4 x^2 -14 x + 6$$ ⇒ $$1 / 2, 3$$ $$y = x^2 -4 x -12$$ ⇒ $$-2, 6$$ $$y = 4 x^2 -10 x -6$$ ⇒ $$-1 / 2, 3$$ $$y = 3 x^2 + 3 x -6$$ ⇒ $$1, -2$$ $$y = x^2 + 7 x + 12$$ ⇒ $$-4, -3$$ $$y = x^2 + 4 x -5$$ ⇒ $$-5, 1$$ $$y = 2 x^2 -5 x -3$$ ⇒ $$-1 / 2, 3$$ $$y = 3 x^2 + 8 x -3$$ ⇒ $$1 / 3, -3$$ $$y = 4 x^2 -9 x -9$$ ⇒ $$-3 / 4, 3$$ $$y = x^2 + 8 x + 12$$ ⇒ $$-6, -2$$