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3w^2-5w-28=0
a = 3; b = -5; c = -28;
Δ = b2-4ac
Δ = -52-4·3·(-28)
Δ = 361
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{361}=19$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*3}=\frac{-14}{6} =-2+1/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*3}=\frac{24}{6} =4 $
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