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q^2-28q-60=0
a = 1; b = -28; c = -60;
Δ = b2-4ac
Δ = -282-4·1·(-60)
Δ = 1024
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{1024}=32$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-32}{2*1}=\frac{-4}{2} =-2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+32}{2*1}=\frac{60}{2} =30 $
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