q^2+23q-24=0

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Solution for q^2+23q-24=0 equation:



q^2+23q-24=0
a = 1; b = 23; c = -24;
Δ = b2-4ac
Δ = 232-4·1·(-24)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*1}=\frac{-48}{2} =-24 $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*1}=\frac{2}{2} =1 $

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