t2-9t+14=0

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Solution for t2-9t+14=0 equation:



t2-9t+14=0
We add all the numbers together, and all the variables
t^2-9t+14=0
a = 1; b = -9; c = +14;
Δ = b2-4ac
Δ = -92-4·1·14
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*1}=\frac{4}{2} =2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*1}=\frac{14}{2} =7 $

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