(8x+1)(x+3)=0

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Solution for (8x+1)(x+3)=0 equation:



(8x+1)(x+3)=0
We multiply parentheses ..
(+8x^2+24x+x+3)=0
We get rid of parentheses
8x^2+24x+x+3=0
We add all the numbers together, and all the variables
8x^2+25x+3=0
a = 8; b = 25; c = +3;
Δ = b2-4ac
Δ = 252-4·8·3
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*8}=\frac{-48}{16} =-3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*8}=\frac{-2}{16} =-1/8 $

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