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(5y-6)(4y+7)=0
We multiply parentheses ..
(+20y^2+35y-24y-42)=0
We get rid of parentheses
20y^2+35y-24y-42=0
We add all the numbers together, and all the variables
20y^2+11y-42=0
a = 20; b = 11; c = -42;
Δ = b2-4ac
Δ = 112-4·20·(-42)
Δ = 3481
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{3481}=59$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-59}{2*20}=\frac{-70}{40} =-1+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+59}{2*20}=\frac{48}{40} =1+1/5 $
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