Looking at the graph we can see that the roots occur at (-1, 0) and (3, 0), therefore \(x = - 1\) and \(x = 3\). Now try the example question below. Solve the quadratic equation \({x^2} ...

The quadratic formula for a quadratic ... to find the square root of a negative number, so the equation has no real solutions. The graph of \(y = x^2 + 2x + 5\) does not cross or touch the x ...