3r^2+8r-80=0

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Solution for 3r^2+8r-80=0 equation:



3r^2+8r-80=0
a = 3; b = 8; c = -80;
Δ = b2-4ac
Δ = 82-4·3·(-80)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-32}{2*3}=\frac{-40}{6} =-6+2/3 $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+32}{2*3}=\frac{24}{6} =4 $

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