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2w^2+3w-44=0
a = 2; b = 3; c = -44;
Δ = b2-4ac
Δ = 32-4·2·(-44)
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$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-19}{2*2}=\frac{-22}{4} =-5+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+19}{2*2}=\frac{16}{4} =4 $
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