20+3*x-4.9*x^2=0

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Solution for 20+3*x-4.9*x^2=0 equation: 



20+3x-4.9x^2=0

a = -4.9; b = 3; c = +20;
Δ = b2-4ac
Δ = 32-4·(-4.9)·20
Δ = 401
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{-(3)-\sqrt{401}}{2*-4.9}=\frac{-3-\sqrt{401}}{-9.8}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{401}}{2*-4.9}=\frac{-3+\sqrt{401}}{-9.8}
$

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