Please someone help me, I don't get it

Answer:
a) x = 1.5 and x = -0.3
b) x = -8 and x = 5
Step-by-step explanation:
a)
The given equation follows the general structure: ax² + bx + c = 0.
Therefore, if a = 5, b = -6, and c = -2, you can substitute the values into the quadratic formula and solve for "x".
b)
Another way of solving polynomials is through factorization. After rearranging the equation to fit the general structure of a quadratic (as seen above), you can factor by asking yourself the question, which 2 numbers multiply to "c" (-40) and add to "b" (3)? The answers will make up your factors.
[tex]\huge\textsf{Hey there!}[/tex]
[tex]\huge\textbf{Equation \#1. }[/tex]
[tex]\mathsf{5x^2 - 6x - 2 = 0}[/tex]
[tex]\huge\textbf{Use the quadratic formula to solve:}[/tex]
[tex]\mathsf{x = \dfrac{-(-6)\pm \sqrt{(-6)^2 - 4(5)(-2)}}{2(5)}}[/tex]
[tex]\huge\textbf{Simplify it: }[/tex]
[tex]\mathsf{x = \dfrac{6 \pm \sqrt{76}}{10}}[/tex]
[tex]\huge\textbf{Simplify that as well:}[/tex]
[tex]\mathsf{x = \dfrac{3}{5} + \dfrac{1}{5}\sqrt{19}\ or\ x = \dfrac{3}{5} + (-\dfrac{1}{5})\sqrt{19}}[/tex]
[tex]\huge\textbf{Therefore, your answer should be:}[/tex]
[tex]\huge\boxed{\mathsf{x \approx 1.5 \ or\ x\approx -0.3{}\ }}\huge\checkmark[/tex]
[tex]\huge\textbf{Equation \#2.}[/tex]
[tex]\mathsf{x^2 + 3x = 40}[/tex]
[tex]\huge\textbf{Subtract 40 to both sides:}[/tex]
[tex]\mathsf{x^2 + 3x - 40 = 40 - 40}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\mathsf{x^2+ 3x - 40 = 0}[/tex]
[tex]\huge\textbf{Factor the left side of the equation:}[/tex]
[tex]\mathsf{(x - 5)\times (x + 8) = 0}[/tex]
[tex]\mathsf{(x - 5)(x + 8) = 0}[/tex]
[tex]\huge\textbf{Set the factors to equal to 0:}[/tex]
[tex]\mathsf{x - 5 = 0 \ or\ even\ x + 8 = 0}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\mathsf{x = 5\ or\ x = -8}[/tex]
[tex]\huge\textbf{Therefore, your answer should be:}[/tex]
[tex]\huge\boxed{\mathsf{x = 5\ or \ x = -8}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]