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3w^2+35w-12=0
a = 3; b = 35; c = -12;
Δ = b2-4ac
Δ = 352-4·3·(-12)
Δ = 1369
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{1369}=37$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-37}{2*3}=\frac{-72}{6} =-12 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+37}{2*3}=\frac{2}{6} =1/3 $
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