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-5t^2-2t+36=12
We move all terms to the left:
-5t^2-2t+36-(12)=0
We add all the numbers together, and all the variables
-5t^2-2t+24=0
a = -5; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-5)·24
Δ = 484
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{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-5}=\frac{-20}{-10} =+2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-5}=\frac{24}{-10} =-2+2/5 $
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