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1.2+20x-4.9x^2=0
a = -4.9; b = 20; c = +1.2;
Δ = b2-4ac
Δ = 202-4·(-4.9)·1.2
Δ = 423.52
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{-(20)-\sqrt{423.52}}{2*-4.9}=\frac{-20-\sqrt{423.52}}{-9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{423.52}}{2*-4.9}=\frac{-20+\sqrt{423.52}}{-9.8} $
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