3y2+25y-18=0

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Solution for 3y2+25y-18=0 equation:



3y^2+25y-18=0
a = 3; b = 25; c = -18;
Δ = b2-4ac
Δ = 252-4·3·(-18)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-29}{2*3}=\frac{-54}{6} =-9 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+29}{2*3}=\frac{4}{6} =2/3 $

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