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10t^2+25t-90=0
a = 10; b = 25; c = -90;
Δ = b2-4ac
Δ = 252-4·10·(-90)
Δ = 4225
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{4225}=65$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-65}{2*10}=\frac{-90}{20} =-4+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+65}{2*10}=\frac{40}{20} =2 $
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