Hi there!
The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation.
x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}
x = \frac{-5 \pm \sqrt{25 - 12} }{2}
x = \frac{-5 \pm \sqrt{13} }{2}
x = \frac{-5 \pm 3.61 }{2}
x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}
x=-0.695 \ \textgreater \ \ \textgreater \ -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31
Therefore, the solutions to the quadratic equation are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!
Your answer is 4.31
