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4t^2+30t-100=0
a = 4; b = 30; c = -100;
Δ = b2-4ac
Δ = 302-4·4·(-100)
Δ = 2500
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{2500}=50$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-50}{2*4}=\frac{-80}{8} =-10 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+50}{2*4}=\frac{20}{8} =2+1/2 $
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