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X^2+6X-528=0
a = 1; b = 6; c = -528;
Δ = b2-4ac
Δ = 62-4·1·(-528)
Δ = 2148
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2148}=\sqrt{4*537}=\sqrt{4}*\sqrt{537}=2\sqrt{537}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{537}}{2*1}=\frac{-6-2\sqrt{537}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{537}}{2*1}=\frac{-6+2\sqrt{537}}{2} $
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