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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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t  - 5*t - 14 = 0
(t2−5t)−14=0\left(t^{2} - 5 t\right) - 14 = 0
Simplify
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=D−b2at_{1} = \frac{\sqrt{D} - b}{2 a}
t2=−D−b2at_{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=7t_{1} = 7
t2=−2t_{2} = -2
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
05-15-10-510152025-250250
Plot the graph f1(x)
Plot the graph f2(x)
Rapid solution [src] 
t1 = -2
t1=−2t_{1} = -2 
t2 = 7
t2=7t_{2} = 7 
t2 = 7
Sum and product of roots [src] 
sum
-2 + 7
−2+7-2 + 7 
=
5
55 
product
-2*7
−14- 14 
=
-14
−14-14 
-14
Numerical answer [src] 
t1 = -2.0
t2 = 7.0
t2 = 7.0
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