5r^2+7r-1=0

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Solution for 5r^2+7r-1=0 equation: 



5r^2+7r-1=0

a = 5; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·5·(-1)
Δ = 69
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{-(7)-\sqrt{69}}{2*5}=\frac{-7-\sqrt{69}}{10}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{69}}{2*5}=\frac{-7+\sqrt{69}}{10}
$

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