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5s^2-48s-20=0
a = 5; b = -48; c = -20;
Δ = b2-4ac
Δ = -482-4·5·(-20)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2704}=52$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-52}{2*5}=\frac{-4}{10} =-2/5 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+52}{2*5}=\frac{100}{10} =10 $
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