3x^2+x-8=1

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



3x^2+x-8=1
We move all terms to the left:
3x^2+x-8-(1)=0
We add all the numbers together, and all the variables
3x^2+x-9=0
a = 3; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·3·(-9)
Δ = 109
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{-(1)-\sqrt{109}}{2*3}=\frac{-1-\sqrt{109}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{109}}{2*3}=\frac{-1+\sqrt{109}}{6} $

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