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x^2+9x=270
We move all terms to the left:
x^2+9x-(270)=0
a = 1; b = 9; c = -270;
Δ = b2-4ac
Δ = 92-4·1·(-270)
Δ = 1161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{129}}{2*1}=\frac{-9-3\sqrt{129}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{129}}{2*1}=\frac{-9+3\sqrt{129}}{2} $
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