4r^2+21r-182=0

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Solution for 4r^2+21r-182=0 equation:



4r^2+21r-182=0
a = 4; b = 21; c = -182;
Δ = b2-4ac
Δ = 212-4·4·(-182)
Δ = 3353
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{3353}}{2*4}=\frac{-21-\sqrt{3353}}{8} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{3353}}{2*4}=\frac{-21+\sqrt{3353}}{8} $

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