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7y^2+3y-10=0
a = 7; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·7·(-10)
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*7}=\frac{-20}{14} =-1+3/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*7}=\frac{14}{14} =1 $
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