(4x+1)(5x+2)=0

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Solution for (4x+1)(5x+2)=0 equation:



(4x+1)(5x+2)=0
We multiply parentheses ..
(+20x^2+8x+5x+2)=0
We get rid of parentheses
20x^2+8x+5x+2=0
We add all the numbers together, and all the variables
20x^2+13x+2=0
a = 20; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·20·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*20}=\frac{-16}{40} =-2/5 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*20}=\frac{-10}{40} =-1/4 $

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