-0,5q^2+6q+12,5=0

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Solution for -0,5q^2+6q+12,5=0 equation:



-0.5q^2+6q+12.5=0
a = -0.5; b = 6; c = +12.5;
Δ = b2-4ac
Δ = 62-4·(-0.5)·12.5
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-\sqrt{61}}{2*-0.5}=\frac{-6-\sqrt{61}}{-1} $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+\sqrt{61}}{2*-0.5}=\frac{-6+\sqrt{61}}{-1} $

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