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7x(1/7+2)=10
We move all terms to the left:
7x(1/7+2)-(10)=0
We multiply parentheses
7x^2+14x-10=0
a = 7; b = 14; c = -10;
Δ = b2-4ac
Δ = 142-4·7·(-10)
Δ = 476
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{119}}{2*7}=\frac{-14-2\sqrt{119}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{119}}{2*7}=\frac{-14+2\sqrt{119}}{14} $
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