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2t^2-8t=6
We move all terms to the left:
2t^2-8t-(6)=0
a = 2; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·2·(-6)
Δ = 112
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{7}}{2*2}=\frac{8-4\sqrt{7}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*2}=\frac{8+4\sqrt{7}}{4} $
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