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r2+r=15
We move all terms to the left:
r2+r-(15)=0
We add all the numbers together, and all the variables
r^2+r-15=0
a = 1; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·1·(-15)
Δ = 61
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{61}}{2*1}=\frac{-1-\sqrt{61}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{61}}{2*1}=\frac{-1+\sqrt{61}}{2} $
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