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242=3x^2+x-2
We move all terms to the left:
242-(3x^2+x-2)=0
We get rid of parentheses
-3x^2-x+2+242=0
We add all the numbers together, and all the variables
-3x^2-1x+244=0
a = -3; b = -1; c = +244;
Δ = b2-4ac
Δ = -12-4·(-3)·244
Δ = 2929
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{2929}}{2*-3}=\frac{1-\sqrt{2929}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{2929}}{2*-3}=\frac{1+\sqrt{2929}}{-6} $
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