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2a^2-9a+10=0
a = 2; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·2·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*2}=\frac{8}{4} =2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*2}=\frac{10}{4} =2+1/2 $
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