3x^2+6x-8=4x

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Solution for 3x^2+6x-8=4x equation:



3x^2+6x-8=4x
We move all terms to the left:
3x^2+6x-8-(4x)=0
We add all the numbers together, and all the variables
3x^2+2x-8=0
a = 3; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·3·(-8)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*3}=\frac{-12}{6} =-2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*3}=\frac{8}{6} =1+1/3 $

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