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7x^2+2x-16=0
a = 7; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·7·(-16)
Δ = 452
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{113}}{2*7}=\frac{-2-2\sqrt{113}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{113}}{2*7}=\frac{-2+2\sqrt{113}}{14} $
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