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21x^2+29x-10=0
a = 21; b = 29; c = -10;
Δ = b2-4ac
Δ = 292-4·21·(-10)
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*21}=\frac{-70}{42} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*21}=\frac{12}{42} =2/7 $
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