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0=2t^2+8t+6
We move all terms to the left:
0-(2t^2+8t+6)=0
We add all the numbers together, and all the variables
-(2t^2+8t+6)=0
We get rid of parentheses
-2t^2-8t-6=0
a = -2; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·(-2)·(-6)
Δ = 16
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*-2}=\frac{4}{-4} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*-2}=\frac{12}{-4} =-3 $
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