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(4+y)(3y-7)=0
We add all the numbers together, and all the variables
(y+4)(3y-7)=0
We multiply parentheses ..
(+3y^2-7y+12y-28)=0
We get rid of parentheses
3y^2-7y+12y-28=0
We add all the numbers together, and all the variables
3y^2+5y-28=0
a = 3; b = 5; c = -28;
Δ = b2-4ac
Δ = 52-4·3·(-28)
Δ = 361
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{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-19}{2*3}=\frac{-24}{6} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+19}{2*3}=\frac{14}{6} =2+1/3 $
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