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a^2+21a-1024=0
a = 1; b = 21; c = -1024;
Δ = b2-4ac
Δ = 212-4·1·(-1024)
Δ = 4537
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{4537}}{2*1}=\frac{-21-\sqrt{4537}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{4537}}{2*1}=\frac{-21+\sqrt{4537}}{2} $
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