(7t+1)(t-2)=0

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Solution for (7t+1)(t-2)=0 equation:



(7t+1)(t-2)=0
We multiply parentheses ..
(+7t^2-14t+t-2)=0
We get rid of parentheses
7t^2-14t+t-2=0
We add all the numbers together, and all the variables
7t^2-13t-2=0
a = 7; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·7·(-2)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-15}{2*7}=\frac{-2}{14} =-1/7 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+15}{2*7}=\frac{28}{14} =2 $

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