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7x^2-25x+4=0
a = 7; b = -25; c = +4;
Δ = b2-4ac
Δ = -252-4·7·4
Δ = 513
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{57}}{2*7}=\frac{25-3\sqrt{57}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{57}}{2*7}=\frac{25+3\sqrt{57}}{14} $
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