(8x-5)(5x-3)=0

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Solution for (8x-5)(5x-3)=0 equation:



(8x-5)(5x-3)=0
We multiply parentheses ..
(+40x^2-24x-25x+15)=0
We get rid of parentheses
40x^2-24x-25x+15=0
We add all the numbers together, and all the variables
40x^2-49x+15=0
a = 40; b = -49; c = +15;
Δ = b2-4ac
Δ = -492-4·40·15
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-1}{2*40}=\frac{48}{80} =3/5 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+1}{2*40}=\frac{50}{80} =5/8 $

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