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(7y-4)(y+5)=28
We move all terms to the left:
(7y-4)(y+5)-(28)=0
We multiply parentheses ..
(+7y^2+35y-4y-20)-28=0
We get rid of parentheses
7y^2+35y-4y-20-28=0
We add all the numbers together, and all the variables
7y^2+31y-48=0
a = 7; b = 31; c = -48;
Δ = b2-4ac
Δ = 312-4·7·(-48)
Δ = 2305
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-\sqrt{2305}}{2*7}=\frac{-31-\sqrt{2305}}{14} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{2305}}{2*7}=\frac{-31+\sqrt{2305}}{14} $
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