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7x^2-42x+1=0
a = 7; b = -42; c = +1;
Δ = b2-4ac
Δ = -422-4·7·1
Δ = 1736
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1736}=\sqrt{4*434}=\sqrt{4}*\sqrt{434}=2\sqrt{434}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{434}}{2*7}=\frac{42-2\sqrt{434}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{434}}{2*7}=\frac{42+2\sqrt{434}}{14} $
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