t^2-2t-7=0

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Solution for t^2-2t-7=0 equation:



t^2-2t-7=0
a = 1; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·1·(-7)
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{-(-2)-4\sqrt{2}}{2*1}=\frac{2-4\sqrt{2}}{2} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{2}}{2*1}=\frac{2+4\sqrt{2}}{2} $

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