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7y^2+10y+3=0
a = 7; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·7·3
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*7}=\frac{-14}{14} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*7}=\frac{-6}{14} =-3/7 $
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