Certainly! Let's solve the quadratic equation [tex]\(2x^2 + 5x + 5 = 0\)[/tex] step-by-step using the quadratic formula.
The general form of a quadratic equation is:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
In this case, we have:
[tex]\[ a = 2, b = 5, \text{and } c = 5 \][/tex]
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Next, we need to find the discriminant (the part under the square root):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = 5^2 - 4(2)(5) = 25 - 40 = -15 \][/tex]
Since the discriminant is negative ([tex]\(\Delta < 0\)[/tex]), the solutions will be complex numbers. We will have to use the imaginary unit [tex]\(i\)[/tex] to express the square root of a negative number.
The solutions for the quadratic equation are:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x = \frac{-5 \pm \sqrt{-15}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-5 \pm \sqrt{15}i}{4} \][/tex]
This simplifies to two solutions:
[tex]\[ x_1 = \frac{-5 + \sqrt{15}i}{4} \][/tex]
[tex]\[ x_2 = \frac{-5 - \sqrt{15}i}{4} \][/tex]
Thus, the solutions to the equation [tex]\(2x^2 + 5x + 5 = 0\)[/tex] are:
[tex]\[ \left(\frac{-5 + \sqrt{15}i}{4}, \frac{-5 - \sqrt{15}i}{4}\right) \][/tex]
In summary, the solution set is:
[tex]\[ x = \frac{-5 + \sqrt{15}i}{4}, \frac{-5 - \sqrt{15}i}{4} \][/tex]
