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6x^2+26x-10=0
a = 6; b = 26; c = -10;
Δ = b2-4ac
Δ = 262-4·6·(-10)
Δ = 916
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{916}=\sqrt{4*229}=\sqrt{4}*\sqrt{229}=2\sqrt{229}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{229}}{2*6}=\frac{-26-2\sqrt{229}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{229}}{2*6}=\frac{-26+2\sqrt{229}}{12} $
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