(3a+1)(2a-5)=0

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Solution for (3a+1)(2a-5)=0 equation:



(3a+1)(2a-5)=0
We multiply parentheses ..
(+6a^2-15a+2a-5)=0
We get rid of parentheses
6a^2-15a+2a-5=0
We add all the numbers together, and all the variables
6a^2-13a-5=0
a = 6; b = -13; c = -5;
Δ = b2-4ac
Δ = -132-4·6·(-5)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*6}=\frac{-4}{12} =-1/3 $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*6}=\frac{30}{12} =2+1/2 $

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