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3X^2+X-252=0
a = 3; b = 1; c = -252;
Δ = b2-4ac
Δ = 12-4·3·(-252)
Δ = 3025
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{3025}=55$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-55}{2*3}=\frac{-56}{6} =-9+1/3 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+55}{2*3}=\frac{54}{6} =9 $
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