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a^2-5a-21=0
a = 1; b = -5; c = -21;
Δ = b2-4ac
Δ = -52-4·1·(-21)
Δ = 109
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{-(-5)-\sqrt{109}}{2*1}=\frac{5-\sqrt{109}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{109}}{2*1}=\frac{5+\sqrt{109}}{2} $
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