7r^2+r-19=0

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Solution for 7r^2+r-19=0 equation:



7r^2+r-19=0
a = 7; b = 1; c = -19;
Δ = b2-4ac
Δ = 12-4·7·(-19)
Δ = 533
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{-(1)-\sqrt{533}}{2*7}=\frac{-1-\sqrt{533}}{14} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{533}}{2*7}=\frac{-1+\sqrt{533}}{14} $

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