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4y^2+19y+21=0
a = 4; b = 19; c = +21;
Δ = b2-4ac
Δ = 192-4·4·21
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*4}=\frac{-24}{8} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*4}=\frac{-14}{8} =-1+3/4 $
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