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63r^2+72r=0
a = 63; b = 72; c = 0;
Δ = b2-4ac
Δ = 722-4·63·0
Δ = 5184
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{5184}=72$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-72}{2*63}=\frac{-144}{126} =-1+1/7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+72}{2*63}=\frac{0}{126} =0 $
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