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4x^2-41x+85=0
a = 4; b = -41; c = +85;
Δ = b2-4ac
Δ = -412-4·4·85
Δ = 321
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-\sqrt{321}}{2*4}=\frac{41-\sqrt{321}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{321}}{2*4}=\frac{41+\sqrt{321}}{8} $
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