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3b^2+26b-9=0
a = 3; b = 26; c = -9;
Δ = b2-4ac
Δ = 262-4·3·(-9)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{784}=28$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-28}{2*3}=\frac{-54}{6} =-9 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+28}{2*3}=\frac{2}{6} =1/3 $
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