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5x^2-36x+7=0
a = 5; b = -36; c = +7;
Δ = b2-4ac
Δ = -362-4·5·7
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-34}{2*5}=\frac{2}{10} =1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+34}{2*5}=\frac{70}{10} =7 $
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