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4.9x^2-20x+15=0
a = 4.9; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·4.9·15
Δ = 106
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{106}}{2*4.9}=\frac{20-\sqrt{106}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{106}}{2*4.9}=\frac{20+\sqrt{106}}{9.8} $
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