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r2+6r=5
We move all terms to the left:
r2+6r-(5)=0
We add all the numbers together, and all the variables
r^2+6r-5=0
a = 1; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·1·(-5)
Δ = 56
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{14}}{2*1}=\frac{-6-2\sqrt{14}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{14}}{2*1}=\frac{-6+2\sqrt{14}}{2} $
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