(3x+5)(4x+1)=0

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Solution for (3x+5)(4x+1)=0 equation:



(3x+5)(4x+1)=0
We multiply parentheses ..
(+12x^2+3x+20x+5)=0
We get rid of parentheses
12x^2+3x+20x+5=0
We add all the numbers together, and all the variables
12x^2+23x+5=0
a = 12; b = 23; c = +5;
Δ = b2-4ac
Δ = 232-4·12·5
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*12}=\frac{-40}{24} =-1+2/3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*12}=\frac{-6}{24} =-1/4 $

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