Question

1. \\(3x^2 + 10x + 6 = 0\\)

a = \\(\underline{\quad\quad}\\) \\(\quad\\) b = \\(\underline{\quad\quad}\\) \\(\quad\\) c = \\(\underline{\quad\quad}\\)
\\(x = \frac{-(\quad)\pm\sqrt{(\quad)^2 - 4(\quad)(\quad)}}{2(\quad)}\\)

Explanation:

Step1: Identify quadratic coefficients

For standard quadratic form $ax^2+bx+c=0$, match terms:
$a=3$, $b=10$, $c=6$

Step2: Substitute into quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Substitute values:
$x=\frac{-(10)\pm\sqrt{(10)^2-4(3)(6)}}{2(3)}$

Step3: Simplify discriminant

Calculate discriminant inside square root:
$10^2-4(3)(6)=100-72=28$

Step4: Simplify final roots

Simplify the expression:
$x=\frac{-10\pm\sqrt{28}}{6}=\frac{-10\pm2\sqrt{7}}{6}=\frac{-5\pm\sqrt{7}}{3}$

Answer:

$a=3$, $b=10$, $c=6$
$x=\frac{-(10)\pm\sqrt{(10)^2-4(3)(6)}}{2(3)}$
Final roots: $x=\frac{-5+\sqrt{7}}{3}$ and $x=\frac{-5-\sqrt{7}}{3}$