-2r2+r+7=0

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Solution for -2r2+r+7=0 equation:



-2r^2+r+7=0
a = -2; b = 1; c = +7;
Δ = b2-4ac
Δ = 12-4·(-2)·7
Δ = 57
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{57}}{2*-2}=\frac{-1-\sqrt{57}}{-4} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*-2}=\frac{-1+\sqrt{57}}{-4} $

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