r^2+5r-50=0

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Solution for r^2+5r-50=0 equation: 



r^2+5r-50=0

a = 1; b = 5; c = -50;
Δ = b2-4ac
Δ = 52-4·1·(-50)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*1}=\frac{-20}{2}
=-10
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*1}=\frac{10}{2}
=5
$

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