2r^2+6r-80=0

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Solution for 2r^2+6r-80=0 equation: 



2r^2+6r-80=0

a = 2; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·2·(-80)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-26}{2*2}=\frac{-32}{4}
=-8
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+26}{2*2}=\frac{20}{4}
=5
$

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