solve \( y=2 x^{2}+3 x+1 \)

To solve the equation \( y = 2x^2 + 3x + 1 \) for \( y = 0 \), we need to find the values of \( x \) where the curve intersects the x-axis. This leads us to solve the quadratic equation \( 2x^2 + 3x + 1 = 0 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = 3, \) and \( c = 1 \). Plugging in these values, we calculate the discriminant \( b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \). Since the discriminant is positive, we have two real solutions: \[ x = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} = \frac{-3 \pm 1}{4} \] Thus, the solutions are \( x = -\frac{1}{2} \) and \( x = -1 \). If you're skimming through quadratic equations, beware of the common mistake of forgetting to factor or simplify correctly. Always double-check your calculations on the discriminant; it can sometimes lead you astray! Also, remember that each quadratic can have a maximum of two real roots depending on the discriminant's sign: zero for one root, positive for two distinct roots, and negative for no real solutions. Happy solving!