2r^2+6r-36=0

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Solution for 2r^2+6r-36=0 equation: 



2r^2+6r-36=0

a = 2; b = 6; c = -36;
Δ = b2-4ac
Δ = 62-4·2·(-36)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*2}=\frac{-24}{4}
=-6
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*2}=\frac{12}{4}
=3
$

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