### Video Transcript

Solve π§ squared plus two plus π π§ plus π equals zero.

Here, we have a quadratic equation, but the coefficients here are nonreal. The use of the letter π§ also indicates to us that weβre going to have complex solutions to this equation. And so, unlike quadratic equations with real coefficients, we have far fewer methods that we can use. So, weβre going to use the quadratic formula. Remember, the quadratic formula says that the solutions to a quadratic equation with coefficients π, π, and π are negative π plus or minus the square root of π squared minus four ππ all over two π. π is the coefficient of π§ squared. Itβs one. π is the coefficient of π§. Thatβs two plus π. And π is the constant term. So, here thatβs π.

Substituting each of these into our formula, and we see that π§ is equal to negative two plus π plus or minus the square root of two plus π squared minus four times one times π all over two times one. Letβs distribute the parentheses over here. We have two plus π times two plus π. Multiplying the first term in each expression, we get four. Multiplying the outer terms, two times π gives us two π. Multiplying the inner terms, we get two times π again. And multiplying the last terms, we get π squared. But of course, π squared is negative one. So, this becomes four plus four π minus one, which simplifies to three plus four π.

Distributing the first parentheses, and we get negative two minus π plus or minus the square root of three plus four π minus four π over two. Now, of course, four π minus four π is zero, which is great because weβre just left with the square root of three and we donβt have to worry about the square root of π. This then simplifies to negative two minus π plus or minus the square root of three over two. Separating the real and imaginary parts, we get negative two plus or minus the square root of three over two minus π over two. And so, we have the two solutions to our equation. They are π§ equals negative two plus root three over two minus π over two and negative two minus root three over two minus π over two.