Certainly. To solve the quadratic equation [tex]\(4x^2 + 16x + 3 = 0\)[/tex], we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
For the equation [tex]\(4x^2 + 16x + 3 = 0\)[/tex]:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 16\)[/tex]
- [tex]\(c = 3\)[/tex]
### Step 1: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 16^2 - 4 \cdot 4 \cdot 3 = 256 - 48 = 208
\][/tex]
### Step 2: Compute the solutions using the quadratic formula
Now, substitute [tex]\(\Delta\)[/tex] back into the quadratic formula:
[tex]\[
x = \frac{{-b \pm \sqrt{\Delta}}}{2a}
\][/tex]
For the solution [tex]\(A\)[/tex]:
[tex]\[
A = \frac{{-b - \sqrt{\Delta}}}{2a} = \frac{{-16 - \sqrt{208}}}{2 \cdot 4} = \frac{{-16 - \sqrt{208}}}{8}
\][/tex]
For the solution [tex]\(B\)[/tex]:
[tex]\[
B = \frac{{-b + \sqrt{\Delta}}}{2a} = \frac{{-16 + \sqrt{208}}}{2 \cdot 4} = \frac{{-16 + \sqrt{208}}}{8}
\][/tex]
### Step 3: Simplify the solutions
Calculating the numerical values for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
A \approx \frac{{-16 - 14.422205101855956}}{8} = \frac{{-30.422205101855956}}{8} \approx -3.802775637731995
\][/tex]
[tex]\[
B \approx \frac{{-16 + 14.422205101855956}}{8} = \frac{{-1.577794898144044}}{8} \approx -0.1972243622680055
\][/tex]
### Conclusion
Thus, the two solutions [tex]\(A\)[/tex] and [tex]\(B\)[/tex] for the quadratic equation [tex]\(4x^2 + 16x + 3 = 0\)[/tex] are:
[tex]\[
A \approx -3.802775637731995 \quad \text{and} \quad B \approx -0.1972243622680055
\][/tex]
Therefore, the solutions to the equation [tex]\(4x^2 + 16x + 3 = 0\)[/tex] are:
[tex]\[
A = -3.802775637731995 \quad \text{and} \quad B = -0.1972243622680055
\][/tex]
So, the equation [tex]\(4 x^2 + 16 x + 3 = 0\)[/tex] has two solutions where [tex]\(A = -3.802775637731995\)[/tex] and [tex]\(B = -0.1972243622680055\)[/tex].
