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x^2+6x-223=0
a = 1; b = 6; c = -223;
Δ = b2-4ac
Δ = 62-4·1·(-223)
Δ = 928
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{58}}{2*1}=\frac{-6-4\sqrt{58}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{58}}{2*1}=\frac{-6+4\sqrt{58}}{2} $
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