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0=-5x^2+36x
We move all terms to the left:
0-(-5x^2+36x)=0
We add all the numbers together, and all the variables
-(-5x^2+36x)=0
We get rid of parentheses
5x^2-36x=0
a = 5; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·5·0
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*5}=\frac{0}{10} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*5}=\frac{72}{10} =7+1/5 $
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