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4x^2-41x+105=0
a = 4; b = -41; c = +105;
Δ = b2-4ac
Δ = -412-4·4·105
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-1}{2*4}=\frac{40}{8} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+1}{2*4}=\frac{42}{8} =5+1/4 $
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