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21x^2-41x-15=0
a = 21; b = -41; c = -15;
Δ = b2-4ac
Δ = -412-4·21·(-15)
Δ = 2941
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{2941}}{2*21}=\frac{41-\sqrt{2941}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{2941}}{2*21}=\frac{41+\sqrt{2941}}{42} $
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