3y2+7y-20=0

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Solution for 3y2+7y-20=0 equation:



3y^2+7y-20=0
a = 3; b = 7; c = -20;
Δ = b2-4ac
Δ = 72-4·3·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*3}=\frac{-24}{6} =-4 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*3}=\frac{10}{6} =1+2/3 $

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