Solve. t^2 + 3t - 28 = 0?

t^2 + 3t - 28 = 0
t^2+7t-4t-28=0
t(t+7)-4(t+7)=0
(t+7)(t-4)=0
t=4, t=-7

(t + 7)(t - 4) = 0
t = - 7 , t = 4

t² + 3t - 28 = 0
(t + 7)(t - 4)

t + 7 = 0, t = - 7
t - 4 = 0, t = 4

Answer: t = - 7, 4

Proof (t is - 7):
7² + 3(- 7) = 28
49 - 21 = 28

Proof (t is 4):
4² + 3(4) = 28
16 + 12 = 28

t²+3t - 28 = 0

Right off, I can see that t=4 and t=-7 are solutions. But if an answer doesn't jump out at you, the quadratic formula will work.
   at²+bt+c=0
where
   a=1
   b=3
   c=-28

t = [-b ± √(b²-4ac)] / (2a)
   = [-(3) ± √((3)² - 4(1)(-28))] / 2(1)
   = [-3 ± √(121)] / 2
   = [-3 ± 11] / 2
   = -1.5 ± 5.5

t₁ = -1.5 + 5.5 = 4
t₂ = -1.5 - 5.5 = -7

Hey, most of the people asking the same question and we are answer that we don't no weather you are getting or not so i will show you the method to do.

to find the value of x ,

ax^2 + bx + c = 0

would be factored into:

(mx+p)(nx+q) =0

Same time you can factorize and find the value of x.

where

mn = a,
pq = c, and
pn + mq = b.

I HOPE THIS WILL HELP!!!!!!!!!!!!!!

t^2 + 3t - 28 = 0
(t + 7)(t - 4) = 0

t + 7 = 0
t = -7

t - 4 = 0
t = 4

∴ t = -7 , 4

Factor:
(t-4)(t+7)=0

find (t-4=0) and (t+7=0):

t=4, -7

t^2 + 3t - 28 = 0
(t +7)(t - 4) = 0
t = 4, -7