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4.9x^2+2x-250=0
a = 4.9; b = 2; c = -250;
Δ = b2-4ac
Δ = 22-4·4.9·(-250)
Δ = 4904
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4904}=\sqrt{4*1226}=\sqrt{4}*\sqrt{1226}=2\sqrt{1226}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1226}}{2*4.9}=\frac{-2-2\sqrt{1226}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1226}}{2*4.9}=\frac{-2+2\sqrt{1226}}{9.8} $
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