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(7+y)(5y-4)=0
We add all the numbers together, and all the variables
(y+7)(5y-4)=0
We multiply parentheses ..
(+5y^2-4y+35y-28)=0
We get rid of parentheses
5y^2-4y+35y-28=0
We add all the numbers together, and all the variables
5y^2+31y-28=0
a = 5; b = 31; c = -28;
Δ = b2-4ac
Δ = 312-4·5·(-28)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*5}=\frac{-70}{10} =-7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*5}=\frac{8}{10} =4/5 $
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