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

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



(4x+1)(x+3)=0
We multiply parentheses ..
(+4x^2+12x+x+3)=0
We get rid of parentheses
4x^2+12x+x+3=0
We add all the numbers together, and all the variables
4x^2+13x+3=0
a = 4; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·4·3
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*4}=\frac{-24}{8} =-3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*4}=\frac{-2}{8} =-1/4 $

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