2x^2-3x+1=8

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Solution for 2x^2-3x+1=8 equation:



2x^2-3x+1=8
We move all terms to the left:
2x^2-3x+1-(8)=0
We add all the numbers together, and all the variables
2x^2-3x-7=0
a = 2; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·2·(-7)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{65}}{2*2}=\frac{3-\sqrt{65}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*2}=\frac{3+\sqrt{65}}{4} $

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