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(4.9x^2+19x-40)=0
We get rid of parentheses
4.9x^2+19x-40=0
a = 4.9; b = 19; c = -40;
Δ = b2-4ac
Δ = 192-4·4.9·(-40)
Δ = 1145
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{-(19)-\sqrt{1145}}{2*4.9}=\frac{-19-\sqrt{1145}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1145}}{2*4.9}=\frac{-19+\sqrt{1145}}{9.8} $
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