-0.01x^2+0.6x+5.9=0

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Solution for -0.01x^2+0.6x+5.9=0 equation: 



-0.01x^2+0.6x+5.9=0

a = -0.01; b = 0.6; c = +5.9;
Δ = b2-4ac
Δ = 0.62-4·(-0.01)·5.9
Δ = 0.596
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{-(0.6)-\sqrt{0.596}}{2*-0.01}=\frac{-0.6-\sqrt{0.596}}{-0.02}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{0.596}}{2*-0.01}=\frac{-0.6+\sqrt{0.596}}{-0.02}
$

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