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-4.9x^2+500x+15=0
a = -4.9; b = 500; c = +15;
Δ = b2-4ac
Δ = 5002-4·(-4.9)·15
Δ = 250294
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{-(500)-\sqrt{250294}}{2*-4.9}=\frac{-500-\sqrt{250294}}{-9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(500)+\sqrt{250294}}{2*-4.9}=\frac{-500+\sqrt{250294}}{-9.8} $
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