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4y^2+20y-144=0
a = 4; b = 20; c = -144;
Δ = b2-4ac
Δ = 202-4·4·(-144)
Δ = 2704
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{2704}=52$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-52}{2*4}=\frac{-72}{8} =-9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+52}{2*4}=\frac{32}{8} =4 $
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