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45f^2+49f=0
a = 45; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·45·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*45}=\frac{-98}{90} =-1+4/45 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*45}=\frac{0}{90} =0 $
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