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45x^2+33x=0
a = 45; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·45·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*45}=\frac{-66}{90} =-11/15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*45}=\frac{0}{90} =0 $
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