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x^2+10x-47=0
a = 1; b = 10; c = -47;
Δ = b2-4ac
Δ = 102-4·1·(-47)
Δ = 288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12\sqrt{2}}{2*1}=\frac{-10-12\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12\sqrt{2}}{2*1}=\frac{-10+12\sqrt{2}}{2} $
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