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5x^2=41x-36=0
We move all terms to the left:
5x^2-(41x-36)=0
We get rid of parentheses
5x^2-41x+36=0
a = 5; b = -41; c = +36;
Δ = b2-4ac
Δ = -412-4·5·36
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-31}{2*5}=\frac{10}{10} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+31}{2*5}=\frac{72}{10} =7+1/5 $
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