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4y^2+20y+11=0
a = 4; b = 20; c = +11;
Δ = b2-4ac
Δ = 202-4·4·11
Δ = 224
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{14}}{2*4}=\frac{-20-4\sqrt{14}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{14}}{2*4}=\frac{-20+4\sqrt{14}}{8} $
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