(8x+3)(2x-5)=0

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Solution for (8x+3)(2x-5)=0 equation:



(8x+3)(2x-5)=0
We multiply parentheses ..
(+16x^2-40x+6x-15)=0
We get rid of parentheses
16x^2-40x+6x-15=0
We add all the numbers together, and all the variables
16x^2-34x-15=0
a = 16; b = -34; c = -15;
Δ = b2-4ac
Δ = -342-4·16·(-15)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-46}{2*16}=\frac{-12}{32} =-3/8 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+46}{2*16}=\frac{80}{32} =2+1/2 $

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