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8x^2+26x-11=0
a = 8; b = 26; c = -11;
Δ = b2-4ac
Δ = 262-4·8·(-11)
Δ = 1028
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{257}}{2*8}=\frac{-26-2\sqrt{257}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{257}}{2*8}=\frac{-26+2\sqrt{257}}{16} $
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