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36x^2+15x-9=0
a = 36; b = 15; c = -9;
Δ = b2-4ac
Δ = 152-4·36·(-9)
Δ = 1521
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{1521}=39$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-39}{2*36}=\frac{-54}{72} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+39}{2*36}=\frac{24}{72} =1/3 $
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