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7x^2+40x+26=0
a = 7; b = 40; c = +26;
Δ = b2-4ac
Δ = 402-4·7·26
Δ = 872
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{872}=\sqrt{4*218}=\sqrt{4}*\sqrt{218}=2\sqrt{218}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{218}}{2*7}=\frac{-40-2\sqrt{218}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{218}}{2*7}=\frac{-40+2\sqrt{218}}{14} $
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