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4.9t^2+44t=98
We move all terms to the left:
4.9t^2+44t-(98)=0
a = 4.9; b = 44; c = -98;
Δ = b2-4ac
Δ = 442-4·4.9·(-98)
Δ = 3856.8
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-\sqrt{3856.8}}{2*4.9}=\frac{-44-\sqrt{3856.8}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+\sqrt{3856.8}}{2*4.9}=\frac{-44+\sqrt{3856.8}}{9.8} $
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