-q^2+27q+415.5=0

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Solution for -q^2+27q+415.5=0 equation:



-q^2+27q+415.5=0
We add all the numbers together, and all the variables
-1q^2+27q+415.5=0
a = -1; b = 27; c = +415.5;
Δ = b2-4ac
Δ = 272-4·(-1)·415.5
Δ = 2391
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{-(27)-\sqrt{2391}}{2*-1}=\frac{-27-\sqrt{2391}}{-2} $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{2391}}{2*-1}=\frac{-27+\sqrt{2391}}{-2} $

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