Y=3x^2+30x+69

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Solution for Y=3x^2+30x+69 equation:



=3Y^2+30Y+69
We move all terms to the left:
-(3Y^2+30Y+69)=0
We get rid of parentheses
-3Y^2-30Y-69=0
a = -3; b = -30; c = -69;
Δ = b2-4ac
Δ = -302-4·(-3)·(-69)
Δ = 72
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:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{2}}{2*-3}=\frac{30-6\sqrt{2}}{-6} $
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{2}}{2*-3}=\frac{30+6\sqrt{2}}{-6} $

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