4(t-2)+4=4t(2t-5)

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



4(t-2)+4=4t(2t-5)
We move all terms to the left:
4(t-2)+4-(4t(2t-5))=0
We multiply parentheses
4t-(4t(2t-5))-8+4=0
We calculate terms in parentheses: -(4t(2t-5)), so:
4t(2t-5)
We multiply parentheses
8t^2-20t
Back to the equation:
-(8t^2-20t)
We add all the numbers together, and all the variables
4t-(8t^2-20t)-4=0
We get rid of parentheses
-8t^2+4t+20t-4=0
We add all the numbers together, and all the variables
-8t^2+24t-4=0
a = -8; b = 24; c = -4;
Δ = b2-4ac
Δ = 242-4·(-8)·(-4)
Δ = 448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{7}}{2*-8}=\frac{-24-8\sqrt{7}}{-16} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{7}}{2*-8}=\frac{-24+8\sqrt{7}}{-16} $

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