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9x^2-23x+10=0
a = 9; b = -23; c = +10;
Δ = b2-4ac
Δ = -232-4·9·10
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-13}{2*9}=\frac{10}{18} =5/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+13}{2*9}=\frac{36}{18} =2 $
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