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4t^2-8t=2
We move all terms to the left:
4t^2-8t-(2)=0
a = 4; b = -8; c = -2;
Δ = b2-4ac
Δ = -82-4·4·(-2)
Δ = 96
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{6}}{2*4}=\frac{8-4\sqrt{6}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{6}}{2*4}=\frac{8+4\sqrt{6}}{8} $
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