3x2+8x=6x+1

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Solution for 3x2+8x=6x+1 equation:



3x^2+8x=6x+1
We move all terms to the left:
3x^2+8x-(6x+1)=0
We get rid of parentheses
3x^2+8x-6x-1=0
We add all the numbers together, and all the variables
3x^2+2x-1=0
a = 3; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·3·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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