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7x+7x^2-50=0
a = 7; b = 7; c = -50;
Δ = b2-4ac
Δ = 72-4·7·(-50)
Δ = 1449
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{161}}{2*7}=\frac{-7-3\sqrt{161}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{161}}{2*7}=\frac{-7+3\sqrt{161}}{14} $
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