(8x+1)(7x+4)=0

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Solution for (8x+1)(7x+4)=0 equation:



(8x+1)(7x+4)=0
We multiply parentheses ..
(+56x^2+32x+7x+4)=0
We get rid of parentheses
56x^2+32x+7x+4=0
We add all the numbers together, and all the variables
56x^2+39x+4=0
a = 56; b = 39; c = +4;
Δ = b2-4ac
Δ = 392-4·56·4
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-25}{2*56}=\frac{-64}{112} =-4/7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+25}{2*56}=\frac{-14}{112} =-1/8 $

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