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3w^2+w-154=0
a = 3; b = 1; c = -154;
Δ = b2-4ac
Δ = 12-4·3·(-154)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1849}=43$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-43}{2*3}=\frac{-44}{6} =-7+1/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+43}{2*3}=\frac{42}{6} =7 $
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