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3m^2+26m-8=0
a = 3; b = 26; c = -8;
Δ = b2-4ac
Δ = 262-4·3·(-8)
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{193}}{2*3}=\frac{-26-2\sqrt{193}}{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{193}}{2*3}=\frac{-26+2\sqrt{193}}{6} $
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