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7y^2+10y-48=0
a = 7; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·7·(-48)
Δ = 1444
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{1444}=38$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-38}{2*7}=\frac{-48}{14} =-3+3/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+38}{2*7}=\frac{28}{14} =2 $
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