Question

what are the solutions of the equation $3x^2 + 13x = 10$?
a 5
b 2
c $\frac{2}{3}$
d 0
e $-\frac{2}{3}$
f $-2$
-5

Explanation:

Step1: Rearrange the equation

First, rewrite the equation \(3x^{2}+13x = 10\) in standard quadratic form \(ax^{2}+bx + c = 0\). Subtract 10 from both sides:
\(3x^{2}+13x - 10 = 0\)

Step2: Factor the quadratic (or use quadratic formula)

Let's try factoring. We need two numbers that multiply to \(3\times(-10)=-30\) and add up to 13. The numbers are 15 and -2.
Rewrite the middle term:
\(3x^{2}+15x - 2x - 10 = 0\)
Group the terms:
\((3x^{2}+15x)+(-2x - 10)=0\)
Factor out common factors from each group:
\(3x(x + 5)-2(x + 5)=0\)
Now, factor out \((x + 5)\):
\((3x - 2)(x + 5)=0\)

Step3: Solve for x

Set each factor equal to zero:

1. \(3x - 2 = 0\)

Add 2 to both sides: \(3x = 2\)
Divide by 3: \(x=\frac{2}{3}\)

1. \(x + 5 = 0\)

Subtract 5 from both sides: \(x=-5\)

Answer:

The solutions are \(x = \frac{2}{3}\) (option C) and \(x=-5\) (the blue - boxed option, which is equivalent to the unlabeled blue option with - 5). So the correct options are C. \(\frac{2}{3}\) and the option with - 5 (the blue one, which can be considered as a valid solution here, and also F? Wait, no, the options are A:5, B:2, C:\(\frac{2}{3}\), D:0, E:\(-\frac{2}{3}\), F:-2, and the blue - 5. Wait, maybe a typo in the problem's options, but from our solution, \(x=-5\) and \(x = \frac{2}{3}\). Looking at the options, C is \(\frac{2}{3}\) and the blue one is - 5 (which is not labeled with a letter but is a solution). But among the given labeled options, C (\(\frac{2}{3}\)) and if we consider the blue - 5 as a solution (maybe a mislabeled option, but from our calculation, \(x=-5\) and \(x=\frac{2}{3}\)). So the solutions are \(x=-5\) and \(x = \frac{2}{3}\), so the correct options are the one with - 5 (the blue boxed) and C. \(\frac{2}{3}\)