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t^2+5t-14=0 equation

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Numerical solution:

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The solution

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 2               
t  + 5*t - 14 = 0
(t2+5t)14=0\left(t^{2} + 5 t\right) - 14 = 0
Detail solution
This equation is of the form
a*t^2 + b*t + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
t1=Db2at_{1} = \frac{\sqrt{D} - b}{2 a}
t2=Db2at_{2} = \frac{- \sqrt{D} - b}{2 a}
where D = b^2 - 4*a*c - it is the discriminant.
Because
a=1a = 1
b=5b = 5
c=14c = -14
, then
D = b^2 - 4 * a * c = 

(5)^2 - 4 * (1) * (-14) = 81

Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)

t2 = (-b - sqrt(D)) / (2*a)

or
t1=2t_{1} = 2
t2=7t_{2} = -7
Vieta's Theorem
it is reduced quadratic equation
pt+q+t2=0p t + q + t^{2} = 0
where
p=bap = \frac{b}{a}
p=5p = 5
q=caq = \frac{c}{a}
q=14q = -14
Vieta Formulas
t1+t2=pt_{1} + t_{2} = - p
t1t2=qt_{1} t_{2} = q
t1+t2=5t_{1} + t_{2} = -5
t1t2=14t_{1} t_{2} = -14
The graph
05-25-20-15-10-51015-250250
Sum and product of roots [src]
sum
-7 + 2
7+2-7 + 2
=
-5
5-5
product
-7*2
14- 14
=
-14
14-14
-14
Rapid solution [src]
t1 = -7
t1=7t_{1} = -7
t2 = 2
t2=2t_{2} = 2
t2 = 2
Numerical answer [src]
t1 = -7.0
t2 = 2.0
t2 = 2.0