r^2-6+5r=0

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Solution for r^2-6+5r=0 equation: 



r^2-6+5r=0

a = 1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*1}=\frac{-12}{2}
=-6
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*1}=\frac{2}{2}
=1
$

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