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a^2-4a-9=0
a = 1; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·1·(-9)
Δ = 52
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{13}}{2*1}=\frac{4-2\sqrt{13}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{13}}{2*1}=\frac{4+2\sqrt{13}}{2} $
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