We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Refresh my memory

0
349
1

What is 3t^2-13t=10?

If I remember

3t^2-13t+10=0

(3t+5)(t-2)

3t+5=0  t-2=0

-5 -5 +2 +2

3t=-5     t=2

t=-5/3

My answer came out wrong, but I don't know what I did wrong.  Can someone refresh my memory?

Aug 30, 2017

### Best Answer

#1
0

The formula you are lookin for is:

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

In your case: $$3t^2 - 13t - 10 = 0$$

so: $$a = 3, b = -13, c = -10$$

and you can just calculate x using this coefficients.

Alternatively, you could separate equation using factors (as you tried to do, I think):
$$3t^2 - 13t - 10 = 0 \\ 3t^2 - 15t + 2t - 10 = 0\\ 3t(t-5) + 2(t-5) = 0\\ (t-5)(3t+2)=0\\$$

which yields two solutions:

$$t_1=5 , \\ 3t_2 = -2 \Rightarrow t_2 = -\frac{2}{5}$$

Cheers.

Aug 30, 2017

### 1+0 Answers

#1
0
Best Answer

The formula you are lookin for is:

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

In your case: $$3t^2 - 13t - 10 = 0$$

so: $$a = 3, b = -13, c = -10$$

and you can just calculate x using this coefficients.

Alternatively, you could separate equation using factors (as you tried to do, I think):
$$3t^2 - 13t - 10 = 0 \\ 3t^2 - 15t + 2t - 10 = 0\\ 3t(t-5) + 2(t-5) = 0\\ (t-5)(3t+2)=0\\$$

which yields two solutions:

$$t_1=5 , \\ 3t_2 = -2 \Rightarrow t_2 = -\frac{2}{5}$$

Cheers.

Guest Aug 30, 2017