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3y^2-8+5y=2-3y
We move all terms to the left:
3y^2-8+5y-(2-3y)=0
We add all the numbers together, and all the variables
3y^2+5y-(-3y+2)-8=0
We get rid of parentheses
3y^2+5y+3y-2-8=0
We add all the numbers together, and all the variables
3y^2+8y-10=0
a = 3; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·3·(-10)
Δ = 184
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{46}}{2*3}=\frac{-8-2\sqrt{46}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{46}}{2*3}=\frac{-8+2\sqrt{46}}{6} $
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