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X^2+32X+200=0
a = 1; b = 32; c = +200;
Δ = b2-4ac
Δ = 322-4·1·200
Δ = 224
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{14}}{2*1}=\frac{-32-4\sqrt{14}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{14}}{2*1}=\frac{-32+4\sqrt{14}}{2} $
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