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4f^2+33f=0
a = 4; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·4·0
Δ = 1089
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{1089}=33$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*4}=\frac{-66}{8} =-8+1/4 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*4}=\frac{0}{8} =0 $
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