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23x^2+69x=0
a = 23; b = 69; c = 0;
Δ = b2-4ac
Δ = 692-4·23·0
Δ = 4761
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{4761}=69$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-69}{2*23}=\frac{-138}{46} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+69}{2*23}=\frac{0}{46} =0 $
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