(5b+1)(b+3)=0

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Solution for (5b+1)(b+3)=0 equation:



(5b+1)(b+3)=0
We multiply parentheses ..
(+5b^2+15b+b+3)=0
We get rid of parentheses
5b^2+15b+b+3=0
We add all the numbers together, and all the variables
5b^2+16b+3=0
a = 5; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·5·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*5}=\frac{-30}{10} =-3 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*5}=\frac{-2}{10} =-1/5 $

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