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5x^2=40x
We move all terms to the left:
5x^2-(40x)=0
a = 5; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·5·0
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*5}=\frac{0}{10} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*5}=\frac{80}{10} =8 $
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