3r^2+3r-36=0

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Solution for 3r^2+3r-36=0 equation: 



3r^2+3r-36=0

a = 3; b = 3; c = -36;
Δ = b2-4ac
Δ = 32-4·3·(-36)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*3}=\frac{-24}{6}
=-4
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*3}=\frac{18}{6}
=3
$

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