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21x^2-8x-4=0
a = 21; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·21·(-4)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-20}{2*21}=\frac{-12}{42} =-2/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+20}{2*21}=\frac{28}{42} =2/3 $
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