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4.9x^2+20x-12=0
a = 4.9; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·4.9·(-12)
Δ = 635.2
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{635.2}}{2*4.9}=\frac{-20-\sqrt{635.2}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{635.2}}{2*4.9}=\frac{-20+\sqrt{635.2}}{9.8} $
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