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21x^2-9x-28=0
a = 21; b = -9; c = -28;
Δ = b2-4ac
Δ = -92-4·21·(-28)
Δ = 2433
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{-(-9)-\sqrt{2433}}{2*21}=\frac{9-\sqrt{2433}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{2433}}{2*21}=\frac{9+\sqrt{2433}}{42} $
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