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

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



(a+3)(3a-5)=0
We multiply parentheses ..
(+3a^2-5a+9a-15)=0
We get rid of parentheses
3a^2-5a+9a-15=0
We add all the numbers together, and all the variables
3a^2+4a-15=0
a = 3; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·3·(-15)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*3}=\frac{-18}{6} =-3 $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*3}=\frac{10}{6} =1+2/3 $

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