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x^2+24x-3=0
a = 1; b = 24; c = -3;
Δ = b2-4ac
Δ = 242-4·1·(-3)
Δ = 588
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-14\sqrt{3}}{2*1}=\frac{-24-14\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+14\sqrt{3}}{2*1}=\frac{-24+14\sqrt{3}}{2} $
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