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r(r+3)=42
We move all terms to the left:
r(r+3)-(42)=0
We multiply parentheses
r^2+3r-42=0
a = 1; b = 3; c = -42;
Δ = b2-4ac
Δ = 32-4·1·(-42)
Δ = 177
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{-(3)-\sqrt{177}}{2*1}=\frac{-3-\sqrt{177}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{177}}{2*1}=\frac{-3+\sqrt{177}}{2} $
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