3b2+26b+48=0

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Solution for 3b2+26b+48=0 equation:



3b^2+26b+48=0
a = 3; b = 26; c = +48;
Δ = b2-4ac
Δ = 262-4·3·48
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-10}{2*3}=\frac{-36}{6} =-6 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+10}{2*3}=\frac{-16}{6} =-2+2/3 $

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