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(4y+7)(y-8)=0
We multiply parentheses ..
(+4y^2-32y+7y-56)=0
We get rid of parentheses
4y^2-32y+7y-56=0
We add all the numbers together, and all the variables
4y^2-25y-56=0
a = 4; b = -25; c = -56;
Δ = b2-4ac
Δ = -252-4·4·(-56)
Δ = 1521
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{1521}=39$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-39}{2*4}=\frac{-14}{8} =-1+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+39}{2*4}=\frac{64}{8} =8 $
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