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25r^2-90r+80=0
a = 25; b = -90; c = +80;
Δ = b2-4ac
Δ = -902-4·25·80
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-10}{2*25}=\frac{80}{50} =1+3/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+10}{2*25}=\frac{100}{50} =2 $
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