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(4y-3)(2y+10)=0
We multiply parentheses ..
(+8y^2+40y-6y-30)=0
We get rid of parentheses
8y^2+40y-6y-30=0
We add all the numbers together, and all the variables
8y^2+34y-30=0
a = 8; b = 34; c = -30;
Δ = b2-4ac
Δ = 342-4·8·(-30)
Δ = 2116
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{2116}=46$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-46}{2*8}=\frac{-80}{16} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*8}=\frac{12}{16} =3/4 $
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