-4.9x^2+20x+1.75=0

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Solution for -4.9x^2+20x+1.75=0 equation:



-4.9x^2+20x+1.75=0
a = -4.9; b = 20; c = +1.75;
Δ = b2-4ac
Δ = 202-4·(-4.9)·1.75
Δ = 434.3
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{434.3}}{2*-4.9}=\frac{-20-\sqrt{434.3}}{-9.8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{434.3}}{2*-4.9}=\frac{-20+\sqrt{434.3}}{-9.8} $

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