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36x^2-27x+5=0
a = 36; b = -27; c = +5;
Δ = b2-4ac
Δ = -272-4·36·5
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3}{2*36}=\frac{24}{72} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3}{2*36}=\frac{30}{72} =5/12 $
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