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3x^2+x-444=0
a = 3; b = 1; c = -444;
Δ = b2-4ac
Δ = 12-4·3·(-444)
Δ = 5329
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{5329}=73$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-73}{2*3}=\frac{-74}{6} =-12+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+73}{2*3}=\frac{72}{6} =12 $
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