4q^2+20q+24=0

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Solution for 4q^2+20q+24=0 equation: 



4q^2+20q+24=0

a = 4; b = 20; c = +24;
Δ = b2-4ac
Δ = 202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*4}=\frac{-24}{8}
=-3
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*4}=\frac{-16}{8}
=-2
$

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