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2-2t^2-3t=0
a = -2; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·(-2)·2
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-2}=\frac{-2}{-4} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-2}=\frac{8}{-4} =-2 $
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