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2y^2+y-10=0
a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*2}=\frac{-10}{4} =-2+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4} =2 $
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