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5y^2+28y-49=0
a = 5; b = 28; c = -49;
Δ = b2-4ac
Δ = 282-4·5·(-49)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-42}{2*5}=\frac{-70}{10} =-7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+42}{2*5}=\frac{14}{10} =1+2/5 $
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