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X^2+3X=324
We move all terms to the left:
X^2+3X-(324)=0
a = 1; b = 3; c = -324;
Δ = b2-4ac
Δ = 32-4·1·(-324)
Δ = 1305
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1305}=\sqrt{9*145}=\sqrt{9}*\sqrt{145}=3\sqrt{145}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{145}}{2*1}=\frac{-3-3\sqrt{145}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{145}}{2*1}=\frac{-3+3\sqrt{145}}{2} $
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