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4.9x^2+18x-48=0
a = 4.9; b = 18; c = -48;
Δ = b2-4ac
Δ = 182-4·4.9·(-48)
Δ = 1264.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-\sqrt{1264.8}}{2*4.9}=\frac{-18-\sqrt{1264.8}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+\sqrt{1264.8}}{2*4.9}=\frac{-18+\sqrt{1264.8}}{9.8} $
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