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

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



5(2x+1)(x+2)=0
We multiply parentheses ..
5(+2x^2+4x+x+2)=0
We multiply parentheses
10x^2+20x+5x+10=0
We add all the numbers together, and all the variables
10x^2+25x+10=0
a = 10; b = 25; c = +10;
Δ = b2-4ac
Δ = 252-4·10·10
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*10}=\frac{-40}{20} =-2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*10}=\frac{-10}{20} =-1/2 $

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