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(3y+7)(y-2)=0
We multiply parentheses ..
(+3y^2-6y+7y-14)=0
We get rid of parentheses
3y^2-6y+7y-14=0
We add all the numbers together, and all the variables
3y^2+y-14=0
a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
Δ = 169
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{169}=13$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*3}=\frac{-14}{6} =-2+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6} =2 $
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