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10y^2+21y+2=0
a = 10; b = 21; c = +2;
Δ = b2-4ac
Δ = 212-4·10·2
Δ = 361
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{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*10}=\frac{-40}{20} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*10}=\frac{-2}{20} =-1/10 $
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