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4.9x^2-16x-2.5=0
a = 4.9; b = -16; c = -2.5;
Δ = b2-4ac
Δ = -162-4·4.9·(-2.5)
Δ = 305
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{-(-16)-\sqrt{305}}{2*4.9}=\frac{16-\sqrt{305}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{305}}{2*4.9}=\frac{16+\sqrt{305}}{9.8} $
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