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X^2+20X=468
We move all terms to the left:
X^2+20X-(468)=0
a = 1; b = 20; c = -468;
Δ = b2-4ac
Δ = 202-4·1·(-468)
Δ = 2272
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2272}=\sqrt{16*142}=\sqrt{16}*\sqrt{142}=4\sqrt{142}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{142}}{2*1}=\frac{-20-4\sqrt{142}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{142}}{2*1}=\frac{-20+4\sqrt{142}}{2} $
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