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2x+4.9x^2-48=0
a = 4.9; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·4.9·(-48)
Δ = 944.8
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{-(2)-\sqrt{944.8}}{2*4.9}=\frac{-2-\sqrt{944.8}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{944.8}}{2*4.9}=\frac{-2+\sqrt{944.8}}{9.8} $
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