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q^2-3q-130=0
a = 1; b = -3; c = -130;
Δ = b2-4ac
Δ = -32-4·1·(-130)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-23}{2*1}=\frac{-20}{2} =-10 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+23}{2*1}=\frac{26}{2} =13 $
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