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3a^2+a-30=0
a = 3; b = 1; c = -30;
Δ = b2-4ac
Δ = 12-4·3·(-30)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*3}=\frac{-20}{6} =-3+1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*3}=\frac{18}{6} =3 $
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