Alright! Let's solve the equation [tex]\( x^2 = 18 + 3x \)[/tex] step-by-step to find its roots.
1. Rewrite the equation in standard form: Move all terms to one side of the equation to set it equal to zero.
[tex]\[
x^2 - 3x - 18 = 0
\][/tex]
2. Apply the quadratic formula: Recall that for a quadratic equation of the form [tex]\(Ax^2 + Bx + C = 0\)[/tex], the solutions for [tex]\(x\)[/tex] are given by:
[tex]\[
x = \frac{{-B \pm \sqrt{{B^2 - 4AC}}}}{2A}
\][/tex]
For our equation [tex]\( x^2 - 3x - 18 = 0 \)[/tex]:
- [tex]\(A = 1\)[/tex]
- [tex]\(B = -3\)[/tex]
- [tex]\(C = -18\)[/tex]
3. Calculate the discriminant: The discriminant [tex]\( \Delta \)[/tex] is:
[tex]\[
\Delta = B^2 - 4AC
\][/tex]
Substituting the values:
[tex]\[
\Delta = (-3)^2 - 4(1)(-18) = 9 + 72 = 81
\][/tex]
4. Find the roots: Using the quadratic formula:
[tex]\[
x = \frac{{-(-3) \pm \sqrt{81}}}{2(1)} = \frac{{3 \pm 9}}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{3 + 9}{2} = \frac{12}{2} = 6
\][/tex]
[tex]\[
x = \frac{3 - 9}{2} = \frac{-6}{2} = -3
\][/tex]
So, the roots of the equation [tex]\( x^2 = 18 + 3x \)[/tex] are [tex]\( x = -3 \)[/tex] and [tex]\( x = 6 \)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{x = -3, x = 6}
\][/tex]
