+0

# Refresh my memory

0
310
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

#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

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