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(21x^2)+11x-6=0
a = 21; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·21·(-6)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-25}{2*21}=\frac{-36}{42} =-6/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+25}{2*21}=\frac{14}{42} =1/3 $
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