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

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



(8x-3)(5x+8)=0
We multiply parentheses ..
(+40x^2+64x-15x-24)=0
We get rid of parentheses
40x^2+64x-15x-24=0
We add all the numbers together, and all the variables
40x^2+49x-24=0
a = 40; b = 49; c = -24;
Δ = b2-4ac
Δ = 492-4·40·(-24)
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-79}{2*40}=\frac{-128}{80} =-1+3/5 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+79}{2*40}=\frac{30}{80} =3/8 $

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