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2(10y^2+31y-14)=0
We multiply parentheses
20y^2+62y-28=0
a = 20; b = 62; c = -28;
Δ = b2-4ac
Δ = 622-4·20·(-28)
Δ = 6084
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{6084}=78$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-78}{2*20}=\frac{-140}{40} =-3+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+78}{2*20}=\frac{16}{40} =2/5 $
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