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48a+29a^2=0
a = 29; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·29·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*29}=\frac{-96}{58} =-1+19/29 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*29}=\frac{0}{58} =0 $
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