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4.9x^2+40x-170=0
a = 4.9; b = 40; c = -170;
Δ = b2-4ac
Δ = 402-4·4.9·(-170)
Δ = 4932
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4932}=\sqrt{36*137}=\sqrt{36}*\sqrt{137}=6\sqrt{137}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-6\sqrt{137}}{2*4.9}=\frac{-40-6\sqrt{137}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+6\sqrt{137}}{2*4.9}=\frac{-40+6\sqrt{137}}{9.8} $
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