(.5t-2)(.4t+6)=0

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Solution for (.5t-2)(.4t+6)=0 equation:



(.5t-2)(.4t+6)=0
We multiply parentheses ..
(+t^2+6t-2t-12)=0
We get rid of parentheses
t^2+6t-2t-12=0
We add all the numbers together, and all the variables
t^2+4t-12=0
a = 1; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*1}=\frac{-12}{2} =-6 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*1}=\frac{4}{2} =2 $

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