(3x-8)(x+4)=

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Solution for (3x-8)(x+4)= equation:



(3x-8)(x+4)=
We move all terms to the left:
(3x-8)(x+4)-()=0
We add all the numbers together, and all the variables
(3x-8)(x+4)=0
We multiply parentheses ..
(+3x^2+12x-8x-32)=0
We get rid of parentheses
3x^2+12x-8x-32=0
We add all the numbers together, and all the variables
3x^2+4x-32=0
a = 3; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·3·(-32)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*3}=\frac{-24}{6} =-4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*3}=\frac{16}{6} =2+2/3 $

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