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22x^2-41x+12=0
a = 22; b = -41; c = +12;
Δ = b2-4ac
Δ = -412-4·22·12
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-25}{2*22}=\frac{16}{44} =4/11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+25}{2*22}=\frac{66}{44} =1+1/2 $
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