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3a^2+11a-130=0
a = 3; b = 11; c = -130;
Δ = b2-4ac
Δ = 112-4·3·(-130)
Δ = 1681
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{1681}=41$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-41}{2*3}=\frac{-52}{6} =-8+2/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+41}{2*3}=\frac{30}{6} =5 $
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