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(3y+4)(10-y)=2
We move all terms to the left:
(3y+4)(10-y)-(2)=0
We add all the numbers together, and all the variables
(3y+4)(-1y+10)-2=0
We multiply parentheses ..
(-3y^2+30y-4y+40)-2=0
We get rid of parentheses
-3y^2+30y-4y+40-2=0
We add all the numbers together, and all the variables
-3y^2+26y+38=0
a = -3; b = 26; c = +38;
Δ = b2-4ac
Δ = 262-4·(-3)·38
Δ = 1132
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1132}=\sqrt{4*283}=\sqrt{4}*\sqrt{283}=2\sqrt{283}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{283}}{2*-3}=\frac{-26-2\sqrt{283}}{-6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{283}}{2*-3}=\frac{-26+2\sqrt{283}}{-6} $
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