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2x^2-23x+66=0
a = 2; b = -23; c = +66;
Δ = b2-4ac
Δ = -232-4·2·66
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-1}{2*2}=\frac{22}{4} =5+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+1}{2*2}=\frac{24}{4} =6 $
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