2t^2 + 5t - 3=0?

Question

this is solving quadratic equations by factoring.

Answer 1

t = 0.5 and -3

Answer 2

Ok: 2(t^2) + 5t -3 = 0

Since we are told that we can solve this by factoring we should be able to rewrite this as:

(at+b)(bt+d)=0

Just to point something out, you may not always be able to do this! Sometimes you cannot factor quadratic equations! In that case you need the quadratic formula to find the roots, which you can easily find by searching quadratic formula on the Internet.

So: (at+b)(ct+d)=ac(t^2) +(ad+bc)t +bd
ac=2, (ad+bc)=5 , and bd=-3

let's make a=1, c=2
(d+2b)=5 and bd=-3 so b=3, d=-1

So, putting the values for a,b,c, and d we have (t+3)(2t-1)=0

check: (t+3)(2t-1)=2t^2 -t +6t -3=2t^2 +5t -3, so we are right.

The roots for (t+3)(2t-1)=0 are t=-3 and t=1/2

Answer 3

Since the lead coefficient is 2 the only factors are 1 and 2, the only factors of the constant term are 1, and 3, and since the sign of the constant teerm is negative, look for the factors of 6 that differ by 5.

(2x – 1)(x +3) = 0

Now, by the zero product rule either 2x – 1 = 0 or x + 3 = 0, hence the two solutions are

x = 1/2 or x = –3

Answer 4

let us split 5 into 2 numbers product -6 ( 2* (-3)

5 = 6 -1

now 2t^2+5t- 3
= 2t^2+6t-t-3
= 2t(t+3)-(t+3)
= (2t-1)(t+3)
= 0
means t = 1/2 or t = -3

the above 2 are solutions
I hope I helped

Answer 5

(2t+1)(t-3)=0 which means that t= -1/2 , t= 3

Answer 6

2t^2 + 5t - 3 = 0

(2t -1) (t+3) = 0

Answer 7

(2t - 1)(t + 3) = 0

so 2t -1 = 0 or t + 3 =0

2t - 1 = 0
2t = 1
t = 1/2

t + 3 = 0
t = -3

t = 1/2 or -3

Answer 8

use quadratic equation -b+- sqrt b^2-4(ac)/2a

-5+- sqrt 49/2(2)
-5+-7/4
-5+7/4=1/2
-5-7/4=-3
Your two answers are 1/2 and -3

Answer 9

remember this formula [-b+-sqrt(b^2 - 4ac)]/2a for more complicated equations a=2 b=5 c= -3

Answer 10

(2t- 1)(t+3)

Answer 11

i combined the termes and i got7t to the power3= -3