8t-t^2=8

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Solution for 8t-t^2=8 equation:



8t-t^2=8
We move all terms to the left:
8t-t^2-(8)=0
We add all the numbers together, and all the variables
-1t^2+8t-8=0
a = -1; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·(-1)·(-8)
Δ = 32
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:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2}}{2*-1}=\frac{-8-4\sqrt{2}}{-2} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2}}{2*-1}=\frac{-8+4\sqrt{2}}{-2} $

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