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a^2+3a-36=0
a = 1; b = 3; c = -36;
Δ = b2-4ac
Δ = 32-4·1·(-36)
Δ = 153
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{17}}{2*1}=\frac{-3-3\sqrt{17}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{17}}{2*1}=\frac{-3+3\sqrt{17}}{2} $
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