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2t(t-1)+8=30
We move all terms to the left:
2t(t-1)+8-(30)=0
We add all the numbers together, and all the variables
2t(t-1)-22=0
We multiply parentheses
2t^2-2t-22=0
a = 2; b = -2; c = -22;
Δ = b2-4ac
Δ = -22-4·2·(-22)
Δ = 180
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{5}}{2*2}=\frac{2-6\sqrt{5}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{5}}{2*2}=\frac{2+6\sqrt{5}}{4} $
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