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3x^2+x+24=360
We move all terms to the left:
3x^2+x+24-(360)=0
We add all the numbers together, and all the variables
3x^2+x-336=0
a = 3; b = 1; c = -336;
Δ = b2-4ac
Δ = 12-4·3·(-336)
Δ = 4033
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{4033}}{2*3}=\frac{-1-\sqrt{4033}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{4033}}{2*3}=\frac{-1+\sqrt{4033}}{6} $
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